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maria [59]
3 years ago
12

What distributive property is 4(x+2)=32

Mathematics
2 answers:
Paraphin [41]3 years ago
8 0

Answer:

Answer:

distributive because

Step-by-step explanation:

4 is distributed to x and 2

4x+2=32

Step-by-step explanation:

koban [17]3 years ago
5 0

Answer:

The distributive property lets you multiply a sum by multiplying each addend separately and then add the products. OK, that definition is not really all that helpful for most people. It is easier to understand the meaning if you look at the examples below.

Step-by-step explanation:

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Solve the system of equations displayed in the picture.
bagirrra123 [75]

The answer is D) (-1,-12)

Reason:

1) substitute 2x-10 for y in y=4x-8

2) 2x-10=4x-8

3) -2x-10=-8

4) -2x=2

5) x=-1

6) substitute -1 for x in y=2x-10

7) y=(2)(-1)-10

8) y=-12

Answer

x=-1 and y=-12

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3 years ago
Multiply 37 by 3 now multiply 37 by 6 what would you have to multiply 37 by to get all fives
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You would have to multiply 37 by 15
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2 years ago
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Jaleah needs some string to make bracelets. She needs
rodikova [14]

Answer:

312

Step-by-step explanation:

3/4 × 32 = 24

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288 +24=312

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2 years ago
This is hard my homework says, name three decimals between 16.4 and 16.5. Draw a number line estimating the placement of all fiv
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6 0
3 years ago
Calculate the total heat required to completely turn 20g of ice at -20°c to steam​
PtichkaEL [24]

Answer:

The total heat required to completely turn 20 g of ice at -20°C to steam, is 65880 J

Step-by-step explanation:

The given parameters are;

The mass of the ice = 20 g

The initial temperature of the ice = -20°C

The specific heat capacity of the ice, ΔH(fusion) = 2.03 J/(g·°C)

The latent heat of fusion of ice = 334 J/g

The specific heat capacity of the water = 4.184 J/(g·°C)

The latent heat of evaporation of water = 2501 J/g

The heat required to raise the temperature of the ice to from -20°C to 0°C, Q₁, is given as follows;

Q₁ = 20 × 2.03 × (0 - (-20)) = 812

Q₁ = 812 J

The latent heat required to melt the 20 g of ice at 0°C is given as follows;

Q₂ = ΔH(fusion) × mass, m = 334 × 20 = 6680

Q₂ = 6680 J

The heat, Q₃, required to raise the temperature of the ice to 100°C (boiling point temperature) is given as follows;

Q₃ = 20 × 4.184 × (100 - 0) = 8368

Q₃ = 8368 J

The heat, Q₄, required to evaporate the 20 g of water at 100°C is given as follows

Q₄ = 2501 × 20 = 50020

Q₄ =  50020 J

The total heat, Q_{total} required to completely turn 20 g of ice at -20°C to steam, is given as follows;

Q_{total} = Q₁ + Q₂ + Q₃ + Q₄ =  812 + 6680 + 8368 + 50020 = 65880

Q_{total} = 65880 J.

6 0
3 years ago
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