1. Suppose the length of each side of a square is increased by 5 feet. If the perimeter of

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

Drag each interval to a box to show if the function shown is increasing, decreasing, or neither over that interval.

prisoha [69]

Answer:

Increasing

−5<x<−1

−1<x<1

Decreasing

4<x<7

−8<x<−5

Neither Increasing nor Decreasing

1<x<4

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP WILL MARK BRAINLIEST, SIMPLE MATH PROBLEM

Blizzard [7]

Answer:

12

Step-by-step explanation:

idk but please just mark me brainliest

8 0

3 years ago

EASY POINTS

weeeeeb [17]

Answer:

1 is true

2 is true

3 is true

4 0

2 years ago

Read 2 more answers

How to learn to divide

ElenaW [278]

Answer:

EX. 210/6

Step 1. Set up the equation, 210/6

Step 2. Divide the first digit/number 6 x 30 = 180 210 - 180 = 30

Step 3. rewrite the equation that is left, 30/6 6 x 5 = 30 30 - 30 = 0

Step 4. Only if need be for another equation, repeat step 1 and 2 until you have your answer

The Answer to the equation 210/6 is 35

Try 426/6 by yourself, please don't use a calculator.

4 0

2 years ago