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Lubov Fominskaja [6]
3 years ago
14

PLEASE HELP QUICK Factor out the GCF of 14x4 + 21x2 + 28x

Mathematics
1 answer:
icang [17]3 years ago
6 0

Answer:

14×4+21×2+28=126

Step-by-step explanation:

correct me if I'm wrong

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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
Select the correct answer from each drop-down menu. The table lists the speeds of the fastest roller coasters in North America a
dangina [55]

Answer:

the answer is

6.41

12.78

17.226

Step-by-step explanation:

7 0
3 years ago
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Three vertices of parallelogram WXYZ are X(–2,–3), Y(0, 5), and Z(7, 7). Find the coordinates of vertex W
Serhud [2]

The coordinates of the vertex W are (5 , -1)

Step-by-step explanation:

In the parallelogram, the diagonal bisect each other

To find a missing vertex in a parallelogram do that:

  • Find the mid-point of a diagonal whose endpoints are given
  • Use this mid-point to find the missing vertex
  • The mid point rule is (\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

∵ WXYZ is a parallelogram

∴ Its diagonals are WY and XZ

∵ The diagonal bisect each other

- That mean they have the same mid-point

∴ They intersect each other at their mid-point

∵ x = (-2 , -3) and z = (7 , 7)

∴ x_{1} = -2 and x_{2} = 7

∴ y_{1} = -3 and y_{2} = 7

- Substitute them in the rule of the mid point to find the

   mid-point of XZ

∴ M_{XZ}=(\frac{-2+7}{2},\frac{-3+7}{2})=(2.5 , 2)

∴ The mid-point of diagonals WY and XZ is (2.5 , 2)

Let us use it to find the coordinates of vertex W

∵ W = (x , y) and Y = (0 , 5)

∴ x_{1} = x and x_{2} = 0

∴ y_{1} = y and y_{2} = 5

- Equate 2.5 by the rule of the x-coordinate of the mid-point

∵ 2.5=\frac{x+0}{2}

- Multiply both sides by 2

∴ 5 = x + 0

∴ 5 = x

∴ The x-coordinate of point W is 5

- Equate 2 by the rule of the y-coordinate of the mid-point

∵ 2=\frac{y+5}{2}

- Multiply both sides by 2

∴ 4 = y + 5

- Subtract 5 from both sides

∴ -1 = y

∴ The y-coordinate of point W is -1

The coordinates of the vertex W are (5 , -1)

Learn more:

You can learn more about the mid-point in brainly.com/question/10480770

#LearnwithBrainly

4 0
3 years ago
Eight friends started a business. They will wear either a baseball cap or a shirt imprinted with their logo while working. They
posledela

Answer:The system of equations to describe the situation are

c + y = 8

6c + 3y = 36

Step-by-step explanation:

Let c represent the number of shirts.

Let y represent the number of caps.

Eight friends started a business. They will wear either a baseball cap or a shirt imprinted with their logo while working. This means that

c + y = 8 - - - - - - - - - - -1

They want to spend exactly $36 on the shirts and caps. Shirts cost $6 each and caps cost $3 each. This means that

6c + 3y = 36 - - - - - - - - - - 2

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Answer:

c

Step-by-step explanation:

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