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mote1985 [20]
2 years ago
13

Match each set of numbers with their least common multiple.

Mathematics
1 answer:
mylen [45]2 years ago
5 0

Answer:

27, 12, and 36  = 108

22 and 4  = 44

14 and 12  = 84

25 and 8  =200

9, 8, and 7  = 504

32, 24, and 18  = 288

Step-by-step explanation: You are welcome.

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(2a^2 + ab +2b)+(4a^2-3ab+9)​
maw [93]

Answer:

-2 a b + 2 b + 6 a^2 + 9

Step-by-step explanation:

Simplify the following:

2 a^2 + a b + 2 b + 4 a^2 - 3 a b + 9

Grouping like terms, 2 a^2 + a b + 2 b + 4 a^2 - 3 a b + 9 = (a b - 3 a b) + 2 b + (2 a^2 + 4 a^2) + 9:

(a b - 3 a b) + 2 b + (2 a^2 + 4 a^2) + 9

a b + a b (-3) = -2 a b:

-2 a b + 2 b + (2 a^2 + 4 a^2) + 9

2 a^2 + 4 a^2 = 6 a^2:

Answer:  -2 a b + 2 b + 6 a^2 + 9

5 0
3 years ago
What is 5,570,000 in scientific notation
denpristay [2]
The answer is 5.570x6 all you do is move you decimal point 

7 0
3 years ago
Read 2 more answers
137000 nearest ten thousand
DochEvi [55]
140,000 is your answer. 
5 0
3 years ago
Read 2 more answers
The graph shows the solution to a system of inequalities:
tangare [24]

Answer:

Option A. 4x+3y\leq 12 and x\geq 0

Step-by-step explanation:

we know that

The solution of the first inequality is the shaded area below the solid line 4x+3y=12

The solid line passes through the points (0,4) and (3,0) (the y and x intercepts)

therefore

The first inequality is

4x+3y\leq 12

The solution of the second inequality is the shaded area to the right of the solid line x=0

therefore

The second inequality is

x\geq 0

7 0
3 years ago
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
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