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3 years ago

Graph the set of points. Which model is most appropriate for the set?

1 answer:

6 0

Check the slope between the moving points

slope between (–6, –1) & (–3, 2) = 1 
slope between (-3, 2) & (–1, 4) = 1 
slope between (-1, 4) & (2, 7) = 1 

the points are collinear and make an angle of 45 degrees with the x-axis 

we can have an equation of a line passing through (-6,-1) and slope 1 as 

(y + 1) = 1(x + 6) 
y = x + 5 is your linear model mostly

You might be interested in

Which of the following phrases translates to m - 7?

Sindrei [870]

m is an unknown number that is being decreased by 7.

OPTION A IS WRONG BECAUSE MATHEMATICALLY IT IS WRITTEN AS

$7 - m$

B IS WRONG BECAUSE IT MEANS

$7 - m$

C IS WRONG BECAUSE IT MEANS

$7 - m$

BUT D IS THE ANSWER BECAUSE IT MEANS

$m - 7$

SINCE m IS TREATED AS AN UNKNOWN NUMBER.

GOODLUCK!!!!

6 0

1 year ago

Explain in words how to find -4 times -1.5 on a number line in words

borishaifa [10]

Since both numbers are negative we know that the product will be positive and when something is time 1.5 that is the same as a 50% increase so that is adding half of the original number (4) so the final answer is 6 (4+(half of 4))=6

5 0

3 years ago

The ratio of the ages (in years) of three children is 2:4:5. The sum of their ages is 33. What is the age of each child?

pickupchik [31]

We have 2x + 4x + 5x = 33 so 11x = 33 and x 3.

The children are 6, 12 and 15

5 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