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3 years ago

Pls help me ..........

Mathematics

1 answer:

Natalka [10]3 years ago

5 0

The jogging percentage is 20/50, which is 40%. Total degree of a circle is 360 degrees. You have to calculate 40% of 360 degrees:
0.4*360=144 degrees

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Trong has 12% of each paycheck automatically deposited to his savings account. His last paycheck was for $2,165. How much money

Dovator [93]

18041.66 dollars to Trongs savings account

7 0

3 years ago

The smiths have two children. The sum of their ages is 23. The produce of their ages is 132. How old are the children?

kolezko [41]

For this case we propose a system of equations:

x: Let the variable representing the age of the first child of the Smiths

y: Let the variable representing the age of the second child of the Smiths

According to the data of the statement we have to:

$x + y = 23\\x * y = 132$

From the first equation we have to:

$x = 23-y$

We substitute in the second equation:

$(23-y) * y = 132\\23y-y ^ 2 = 132\\y ^ 2-23y + 132 = 0$

We find the solutions by factoring:

We look for two numbers that, when multiplied, result in 132 and when added, result in 23. These numbers are 11 and 12.

Thus, we have that the factorized equation is:

$(y-11) (y-12) = 0$

Thus, the solutions are: $y_ {1} = 11\\y_ {2} = 12$

So, we can take any of the solutions:

With $y = 11$

Then $x = 23-11 = 12$

Therefore, the ages of the children are 11 and 12 respectively.

Answer:

The ages of the children are 11 and 12 respectively.

6 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

Helpppppppppppp ppppp:/

jeka94

Answer:

Step-by-step explanation:

I hope this helps

3 0

4 years ago

I NEED HELLLLLLLPPPPPPP? SHOW WORK PLSSSSSS IF WORK IS NOT SHOWN I WILL REPORT YOU! SO WATCH OUT!

Phoenix [80]

Answer is D
(1.5,3)

Explanation:

Substitute

1) 3-2(1.5)=0

So 0=0

2) 3=8(1.5)-9

So 3=3

7 0

3 years ago

Pls help me ..........​

Pls help me ..........