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2 years ago

The ratios 9:18, 12:24, 15:30, 18:36, and 21:42 form a pattern of equivalent ratios in the table below.

Mathematics

2 answers:

igor_vitrenko [27]2 years ago

7 0

I don’t think a anyone has responded to this but if they have oh well but my answer is 6:12

Fynjy0 [20]2 years ago

4 0

Answer:

I would say 6:12.

Step-by-step explanation:

it is equivalent to 12:24 because 12/2= 6 and 24/2 = 12 so 6:12 is your answer!

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3x - 2 2x + 1 д B c If x = 5, then the length of BC is

masha68 [24]

So BC is 2x+1 and they are telling you that if your x is 5 then what is BC
So you have to plug in the 5 into your equation
2(5)+1
10+1
11
So BC is 11

4 0

2 years ago

a student finshed 8 of her quiz problems in class. if the ratio of problem she finished to problems she still had left was 4:1 h

andrezito [222]

Answer:

Step-by-step explanation:

I can't really explain this well but, when you double 4 you get 8 (which is how many she did), so since you doubled 4 you also have to double 1 which would be 2. Then add 8 and 2 and you get 10

7 0

2 years ago

Read 2 more answers

40 = 25 + x

RSB [31]

Answer:

True

Step-by-step explanation:

This is true. x + 25 equals 40 in both cases, so they're equal.

4 0

2 years ago

nadya68 [22]

Answer:

Step-by-step explanation:

2, 6, 28, 23,7.64.

because 64 is the furthest away from any other number

5 0

1 year 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

2 years ago