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4 years ago

Is this proportional or non proportional ???

Mathematics

1 answer:

Shtirlitz [24]4 years ago

7 0

Answer:

non-proportional

Step-by-step explanation:

non-proportional because it doesn't pass through the origin.

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Helpanswer question 2

Alona [7]

Answer:

(2,3)

Step-by-step explanation:

-x-(x+1)=-5

-2x-1=-5

-2x=-4

x=2

y=3

5 0

3 years ago

Y=3x + 5 <br> Please help

Evgesh-ka [11]

Answer:

whats x?

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

using distance formula what is the distance point between (-8,4) and (-4,-1) round to nearest tenth

RideAnS [48]

Answer:

6.4 units

Step-by-step explanation:

Distance² = (change in y)² + (change in x)²

Distance² = (4 - -1)² + (-8 - -4)²

Distance² = 25 + 16 = 41

Distance = $\sqrt{41}$ = 6.403124237

The distance is 6.4 units (rounded off to nearest tenth)

8 0

4 years ago

Find the quotient. (12x 2 - 13x - 4) ÷ (4x + 1)

user100 [1]

Answer:

3x-4

Step-by-step explanation:

$(12x^2-13x-4)\div(4x+1)= \\\\(3x-4)(4x+1)\div (4x+1)= \\\\3x-4$

Hope this helps!

8 0

4 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

4 0

4 years ago