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4 years ago

14

Using a graphing utility, find the exact solutions of the system. Round to the nearest hundredth and choose a solution to the sy

stem from the choices below.
y = x^2 + 3x
y = x + 5

a. (2.79, 0.58)
b. (1.79, 8.58)
c. (-0.58, -2.79)
d. (0,4)

1 answer:

Furkat [3]4 years ago

7 0

When I tried it, None of the answers offered are correct. The solutions is where the line/parabola intersect. Tell me if I did anything wrong. https://www.desmos.com/calculator/fpz7tjaugu

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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

4 years ago

Find one integral roots of the polynomial x^3-6x^2+11x-6

lord [1]

Answer:

x = 1

Step-by-step explanation:

Sum the coefficients of the polynomial, that is

1 - 6 + 11 - 6 = 0

Hence x = 1 is a root and (x - 1) is a factor of the polynomial

6 0

3 years ago

A consult needs to make at least $600 this week she earns $120 for each new piece and $60 for each review which of the following

inessss [21]

The consult must make at least 600 dollars. That means the combination of reviews and new written pieces must be at least $600, or greater than or equal to (<span>≥) $600.

You are also told that each new piece = $120 and each review = $60. Let's use variables n = new piece and r = review. (You might be given different variables for your answer choices). That means the total money made from a combination of new pieces and reviews is:
120n (money made from each new piece times n number of new pieces) + 60r (money made from each review times r number of reviews).

Put that all together to get your inequality:
120n + 60r </span>≥ 600

-----

Answer: 120n + 60r ≥ 600

8 0

3 years ago

a recipe cause for 4 cups of flour and 1 1/4 cups of milk how much milk do you use for 2 cups of flour

nignag [31]

Answer:

Step-by-step explanation:

Ratio of flour and milk = 4 : 1 1/4 = 4 : 5/4

4/2 : (5/4)/2 = 2 : 5/4*2 = 2: 5/8

5/8 cup of milk

3 0

4 years ago

Read 2 more answers

Juwan has a third quiz in math on Friday. His two previous quiz grades were 85 and 72. What must he make on the third quiz for h

VLD [36.1K]

Answer:

$x = 107$

Step-by-step explanation:

Given

$Scores = 85\ and\ 72$

$Average = 88$

Required

Determine the third score

Average is calculated using;

$Average = \frac{\sum x}{n}$

In this case:

$n = 3$

$Average = \frac{\sum x}{n}$

Let the third score be x; So:

$Average = \frac{85 + 72 + x}{3}$

Substitute 88 for Average

$88 = \frac{85 + 72 + x}{3}$

$88 = \frac{157 + x}{3}$

Multiply through by 3

$3 * 88 = \frac{157 + x}{3} * 3$

$264 = 157 + x$

$x = 264 - 157$

$x = 107$

4 0

4 years ago