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const2013 [10]
3 years ago
5

72, 12.2. common ratio

Mathematics
1 answer:
Ymorist [56]3 years ago
8 0

This problem is not solvable.

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6x + 7 = 10x then 18x =
Marysya12 [62]

Answer:

31.5 (multiplied it after)

Step-by-step explanation:

I love algebra anyways

The ans is in the picture with the  steps how i got it

(hope this helps can i plz have brainlist :D hehe)

5 0
2 years ago
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A box of chocolates contains five milk chocolates, five dark chocolates, and five white chocolates. You randomly select and eat
Marina CMI [18]

Answer:

it the first choice you ate three chocolate independently

6 0
2 years ago
11
BaLLatris [955]
<h3>Answer:</h3>

x=30

<h3>Solution:</h3>
  • First, use the distributive property:
  • 8x-3(2x-4)=3(x-6)
  • 8x-6x+12=3x-18
  • Simplify:
  • 2x+12=3x-18
  • 2x-3x=-12-18
  • -x=-30
  • x=30

Hope it helps.

Do comment if you have any query.

8 0
2 years ago
A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per lit
Veronika [31]

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = -  (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] =  20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

7 0
3 years ago
I can’t find answers to this question. Please help.
dlinn [17]

Answer:

Did you try looking it up

Step-by-step explanation:

sorry If I am wrong and don't forget to make me brainliest

3 0
3 years ago
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