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Alenkinab [10]
3 years ago
13

The table compares actual lengths in a room to lengths in a blueprint of the room.

Mathematics
1 answer:
krok68 [10]3 years ago
4 0

Answer:

13.5 feets

1 inch

Step-by-step explanation:

Given :

Blueprint length (in.) __2 4 6 8 10 12

Actual length (ft) _____3 6 9 12 15 18

From the data:

Blueprint : actual

2 in : 3 feets

Length of wall on blueprint = 9 inches

Actual length of wall :

2 inches = 3feets

9 inches = x

Cross multiply

2x = 27

x = 27 /2

x = 13.5 feets

B.)

For actual width = 1.5 feets

2 inches = 3feets

x = 1.5 feets

3x = 2 * 1.5

3x = 3

x = 3/3

x = 1 inch

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7(4x - 2) - 4(2x - 8)
vesna_86 [32]

Answer:

20x + 18

Step-by-step explanation:

We need to use the distributive property, where we essentially take the sum of the product of the outside number with each of the inside terms.

In 7(4x - 2), 7 is the outside number and 4x and -2 are the inside numbers, so:

7(4x - 2) = 7 * 4x + 7 * (-2) = 28x - 14

In 4(2x - 8), 4 is the outside number and 2x and -8 are the inside numbers, so:

4(2x - 8) = 4 * 2x + 4 * (-8) = 8x - 32

Now, we have:

28x - 14 - (8x - 32) = 28x - 14 - 8x + 32 = 20x + 18

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3 years ago
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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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