The question is below pls help ASAP

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

Dmitry [639]

At any time $t$ (min), the volume of solution in the tank is

$500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}$

If $A(t)$ is the amount of salt in the tank at any time $t$ , then the solution has a concentration of $\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}$ .

The net rate of change of the amount of salt in the solution, $A'(t)$ , is the difference between the amount flowing in and the amount getting pumped out:

$A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)$

Dropping the units and simplifying, we get the linear ODE

$A'=10+5\sin\dfrac t4-\dfrac A{10}$

$10A'+A=100+50\sin\dfrac t4$

Multiplying both sides by $e^{10t}$ allows us to identify the left side as a derivative of a product:

$10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}$

$\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}$

$e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt$

Integrate and divide both sides by $e^{10t}$ to get

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}$

The tanks starts off with 30 lb of salt, so $A(0)=30$ and we can solve for $C$ to get a particular solution of

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}$

6 0

3 years ago

in the circle, m∠S=33°, mRS=120, and RU is a tangent. the diagram is not drawn to scale. what is m∠U? Please help!

SIZIF [17.4K]

Answer:

27°

Step-by-step explanation:

arc RT = 66

1/2(120 - 66) = 27

8 0

3 years ago

Read 2 more answers

How do yuh evaluate the expression Log 4096 4

Furkat [3]

$log_{4}(4096)$

$4096=4^{6}$

let's replace it

$log_{4}(4^6)$

Them

$log_4(4^6)=6log_{4}(4)$

$log_4(4)=1$

$\boxed{\boxed{log_4(4096)=6}}$

6 0

3 years ago

Read 2 more answers

There is a line through the origin that divides the region bounded by the parabola y=2x-4x^2 and the x-axis into two regions wit

shtirl [24]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

y = 7x - 4x²

7x - 4x² = 0 

x(7 - 4x) = 0 

x = 0, 7/4 

Find the area of the bounded region... 

A = ∫ 7x - 4x² dx |(0 to 7/4) 

A = 7/2 x² - 4/3 x³ |(0 to 7/4) 

A = 7/2(7/4)² - 4/3(7/4)³ - 0 = 3.573 

Half of this area is 1.786, now set up an integral that is equal to this area but bounded by the parabola and the line going through the origin... 

y = mx + c 

c = 0 since it goes through the origin 

The point where the line intersects the parabola we shall call (a, b) 

y = mx ===> b = m(a) 

Slope = m = b/a 

Now we need to integrate from 0 to a to find the area bounded by the parabola and the line... 

1.786 = ∫ 7x - 4x² - (b/a)x dx |(0 to a) 

1.786 = (7/2)x² - (4/3)x³ - (b/2a)x² |(0 to a) 

1.786 = (7/2)a² - (4/3)a³ - (b/2a)a² - 0 

1.786 = (7/2)a² - (4/3)a³ - b(a/2) 

Remember that (a, b) is also a point on the parabola so y = 7x - 4x² ==> b = 7a - 4a² 
Substitute... 

1.786 = (7/2)a² - (4/3)a³ - (7a - 4a²)(a/2) 

1.786 = (7/2)a² - (4/3)a³ - (7/2)a² + 2a³ 

(2/3)a³ = 1.786 

a = ∛[(3/2)(1.786)] 

a = 1.39 

b = 7(1.39) - 4(1.39)² = 2.00 

Slope = m = b/a = 2.00 / 1.39 = 1.44

7 0

3 years ago

Tina lives in a state that charges her 4.5% state income tax on her federal taxable income. If her federal taxable income is $61

Irina-Kira [14]

$ 2772 is paid in state income tax

Solution:

Given that, Tina lives in a state that charges her 4.5% state income tax on her federal taxable income

Federal taxable income = $61,600

To find: Amount paid by Tina in state income tax

Tax rate = 4.5 %

Which means, 4.5 % of $61,600 is paid as tax

$Tax\ Rate = 4.5 \% \text{ of } 61600$

Here, "of" means multiplication or product

$Tax\ Rate = 4.5 \% \times 61600\\\\Tax\ Rate = \frac{4.5}{100} \times 61600\\\\Tax\ Rate = 0.045 \times 61600\\\\Tax\ Rate = 2772$

Thus $ 2772 is paid in state income tax

7 0

3 years ago