Which inequality represents all possible solution of -6n< -12

What is the equation of a line that passes through the point (8, −2) and is parallel to the line whose equation is 3x + 4y = 15?

Dafna1 [17]

Step 1

Find the slope of the given line

we have

$3x+4y=15$

Isolate the variable y

Subtract $3x$ both sides

$3x+4y-3x=15-3x$

$4y=-3x+15$

Divide by $4$ both sides

$y=-\frac{3}{4}x+ \frac{15}{4}$

the slope of the given line is

$m=-\frac{3}{4}$

Step 2

Find the equation of the line that passes through the point $(8,-2)$ and is parallel to the given line

we know that

if two lines are parallel, then their slopes are equal

The equation of the line into point-slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{3}{4}$

$(x1,y1)=(8,-2)$

substitute in the equation

$y+2=-\frac{3}{4}(x-8)$

$y=-\frac{3}{4}x+6-2$

$y=-\frac{3}{4}x+4$ or $3x+4y=16$

therefore

the answer is

$y=-\frac{3}{4}x+4$

or

$3x+4y=16$

8 0

3 years ago

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