What is the ratio of 98:56:84?

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

4 years ago

Clarence works at least 5 hours but not more than 7 hours. he earns $11.60 per hour. the function f(t)=11.6t represents the amou

Pavel [41]

Function: f(t) = 11.6 t

Restrictions: 5 ≤ t ≤ 7

Domain: possible values of t: all real values between 5 and 7 inclusive

Range: possible values of f(t): from 11.6 * 5 to 11.6 *7 = from 58 to 81.2, inclusive.

Answer:

the practical domain is all real numbers from 5 to 7, inclusive. and the practical range is all real numbers from 58 to 81.2

4 0

3 years ago

Read 2 more answers

PLEASE SOLVE THIS :(

UNO [17]

Answer:

cost price $500

Step-by-step explanation:

90/100(x) + 150 = 120/100(x)

(120-90)/100(x) = 150

30/100(x) = 150

30x = 15000

x = 500

3 0

3 years ago

2) 15 was subtracted from a number and that difference was then divided by 0.7. After which, that quotient was multiplied by 3.

Galina-37 [17]

Answer:

A number (x) is 14.3

Step-by-step explanation:

You can write the problems like this and solve starting with z to x

x - 15 = y (15 was subtracted from a number)LAST HERE

x = 15 + -0.7

x = 14.3

y / 0.7 = z (and that difference was then divided by 0.7.)THEN HERE

y = -1 x 0.7

y = -0.7

3z = -3(After which, that quotient was multiplied by 3. The resulting product was -3.) START HERE

z = -3 / 3

z = -1

CHECK

14.3 - 15 = -0.7 -0.7 / 0.7 = -1 3 x -1 = -3

3 0

3 years ago

Giving brainlist !! identify the coefficient, constant and number of terms :)

Serggg [28]

Answer:

The number of terms is like the numbers with the x.

Constant are the numbers without the variable like the number 7 in that equation.

Coefficient numbers are the numbers with the x but not including the x.

Step-by-step explanation:

I think I did it right.

7 0

3 years ago