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3 years ago

Rational numbers with distribute property

Mathematics

1 answer:

Svetradugi [14.3K]3 years ago

4 0

I'm not sure what this means. If you have choices you should list them.

(1/2)*(1/4 + 1/6) is an example of what should be given. There are two ways to solve this.

1. Use the distributive property.

1/2*1/4 + 1/2* 1/6

1/8 + 1/12 Which can be added using the LCD of 24

3/24 + 2/24 = 5/24

Method 2

Add what is inside the brackets first.

1/2 ( 1/4 + 1/6)

1/2(3/12 + 2/12 = 5/12

Now multiply by 1/2

1/2(5/12) = 5*1/(12 * 2) = 5 / 24 Same answer.

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HELP PLEASE!!!!!!! Simplify 6r · s · 4rt. 10rst 10r2st 24r2st

lakkis [162]

The answer would be the third choice.
Here’s why:

First multiply 6 and 4 to get 24.
The multiply r, r, s, and t to get r^2st.
The answer is 24r^2st.

8 0

4 years ago

Read 2 more answers

A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is

tresset_1 [31]

Answer:

a) $x(t) = 13*e^(^-^\frac{t}{100}^)$

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

$\frac{dx}{dt} =$ ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

$conc = \frac{x(t)}{1000}\frac{kg}{L}$

- The amount of salt leaving the tank per unit time can be determined from:

salt flow-out = conc * V( flow-out )

salt flow-out = $\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\$

salt flow-out = $\frac{x(t)}{100}\frac{kg}{min}$

- The ODE becomes:

$\frac{dx}{dt} = 0 - \frac{x}{100}$

- Separate the variables and integrate both sides:

$\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)$

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

$13 = C*e^0 = C$

- The solution to the ODE becomes:

$x(t) = 13*e^(^-^\frac{t}{100}^)$

- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:

$x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg$

- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg

7 0

3 years ago

How you round 348 to the nearest ten and to the nearest hundred explain

givi [52]

Answer:

Nearest ten-350

Nearest hundred-300

Step-by-step explanation:

#When rounding to the nearest ten, we look at the ones digits in a number:

If the ones digit is 0, 1, 2, 3, or 4, you will round down to the previous ten
If the ones digit is 5, 6, 7, 8, or 9, you will round up to the next ten

Given the number 348, 8 in the ones digit greater than 4 so we round to the next ten:

348 ≈350

#When rounding to the nearest hundred, we look at the tens digits in a number:

If the ones digit is less than 5 you will round down to the previous hundred
If the ones digit is greater than 4, you will round up to the next hundred

Given the number 348, 4 in the tens digit less than 5 so we round down to the previous hundred:

348 ≈300

6 0

3 years ago

What is the distance between (-1,4) and (2,0)

Ivahew [28]

Answer: 5

The Distance between the coordinates of two points is measured by distance formula.

Step-by-step explanation:

The Distance between the coordinates of two points P(x1,y1) and Q(x2,y2) is measured by

Distance PQ= $\sqrt{}$ (x2-x1)²+(y2-y1)²

So, the distance between (-1,4 ) and (2,0) can be calculated by

Distance= $\sqrt{}$ (2-(-1))²+(0-4)²

= $\sqrt{}$ (3)²+(-4)²

= $\sqrt{}$ 9+16 = $\sqrt{}$ 25 as $\sqrt{}$ 25 = 5

Distance = 5

So the distance between the two coordinates is 5.

7 0

3 years ago

What's the value of given expression <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B1%20%5Ctimes%201%7D%20" id="TexFormul

arsen [322]

Answer:

Step-by-step explanation:

1st step : multiple the number inside square root

1 x 1 =1

so then think about which time 1 will give you 1

that will 1

6 0

3 years ago

Read 2 more answers