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iVinArrow [24]
3 years ago
8

How to compare 2 5/9 and 21/3

Mathematics
2 answers:
max2010maxim [7]3 years ago
5 0

Answer:

2 5/9 is less than 21/3;  2 5/9 < 21/3

Step-by-step explanation:

<u>Step 1:  Convert 2 5/9 into an improper fraction</u>

2 5/9 = 2 * 9/9 + 5/9 = 18/9 + 5/9 = 23/9

<u>Step 2:  Convert 21/3 so the denominator is 9</u>

21*3 / 3*3 = 63/9

<u>Step 3:  Compare</u>

<em>23/9 < 63/9</em>

<em />

Answer:  2 5/9 is less than 21/3;  2 5/9 < 21/3

Troyanec [42]3 years ago
4 0

Answer:

21/3 > 2 5/9

Step-by-step explanation:

2 × 9 = 18

18 + 5 = 23 which makes the fraction 23/9

23/9 × 3 = 69/27

21/3 × 9 = 189/27

189/27 > 69/27 or 21/3 > 2 5/9

Or Simpler:

21/3 = 7

7 > 2 5/9

So, 21/3 > 2 5/9

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4.2 3x3 + 8x2 + 4x - 7 is not a perfect cube
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Thoughtfully split the expression at hand into groups, each group having two terms :
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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)

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