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3 years ago

Explain how to solve an inequality by finding equivalent inequalities

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

3 0

Answer:

cheat off somebody who has the answer

Step-by-step explanation:

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A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

Substitute: 7.5x = 5.5x + 10

Umnica [9.8K]

The value of x is 5

<h3>What are algebraic expressions?</h3>

Algebraic expressions are expressions made up;

Factors
variables
terms
constants

They also consist of mathematical operations such as addition, multiplication, division, subtraction, parenthesis, brackets, etc

We have the expression as;

7.5x = 5.5x + 10

collect like terms

7.5x - 5.5x = 10

subtract the like terms

2.0x = 10

Make 'x' the subject

x = 10/2. 0

x = 5

Thus, the value of x is 5

Learn more about algebraic expressions here:

brainly.com/question/4344214

#SPJ1

7 0

2 years ago

A marketing firm is asked to estimate the percent of existing customers who would purchase a "digital upgrade" to their basic ca

Ugo [173]

Answer: A. 664

Step-by-step explanation:

Given : A marketing firm is asked to estimate the percent of existing customers who would purchase a "digital upgrade" to their basic cable TV service.

But there is no information regarding the population proportion is mentioned.

Formula to find the samples size , if the prior estimate to the population proportion is unknown :

$n=0.25(\dfrac{z*}{E})^2$

, where E = Margin of error.

z* = Two -tailed critical z-value

We know that critical value for 99% confidence interval = $z*=2.576$ [By z-table]

Margin of error = 0.05

Then, the minimum sample size would become :

$n=0.25(\dfrac{2.576}{0.05})^2$

Simplify,

$n=0.25\times2654.3104=663.5776\approx664$

Thus, the required sample size= 664

Hence, the correct answer is A. 664.

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%2072" id="TexFormula1" title="\sqrt{x} 72" alt="\sqrt{x} 72" align="absmiddle"

AveGali [126]

Answer:

Step-by-step explanation:

√x^72 = √(x^36)^2 = x^36

4 0

3 years ago

The weight of a rock sample is measured to be 2.5 pounds. What is the percent of error in the measurement? 2% 5% 10%

elixir [45]

Answer:

The percent of error in the measurement is 2%

Step-by-step explanation:

The percent of error associated with a reported measurement is calculate using the formula;

$percenterror=\frac{error}{measurement}*100$

The error associated with a measurement is defined as half of the smallest unit of measurement used. The measurement reported was 2.5. The smallest unit of measurement for this reading is 0.1. The error is thus;

error = 0.1/2 = 0.05

The percent of error is thus;

$\frac{0.05}{2.5}*100=2$

6 0

3 years ago