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marissa [1.9K]
3 years ago
11

Solve this problem √-16

Mathematics
1 answer:
a_sh-v [17]3 years ago
3 0

Answer:

4i

Step-by-step explanation:

\sqrt{-1}  = i

Therefore \sqrt{-16} = \sqrt{16} \sqrt{-1} \\=4i

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Which number is a solution of the inequality x <-4? Use the number line to help answer the question.
Fantom [35]

Answer:

0 is the answer

Step-by-step explanation:

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3 years ago
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ycow [4]

Answer:

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eq2 yint= 4

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7 0
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What is the solution to 2(2x +1) = 0.5(2x - 14)?​
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A 150 no container can hold 23.6 g of aluminum what is the density of the aluminum?
Drupady [299]

Answer:

# The density of the aluminum is 59/375 ≅ 0.157 g/cm³

# The volume of the container is 4/17 ≅ 0.235 cm³

# The mass of iron is 1225 g

Step-by-step explanation:

* Lets explain how to solve the problem

- Density is a measurement that compares the amount of matter

 of an object to its volume [ Density = mass/volume]

- Density is found by dividing the mass of an object by its volume

- The unit of measuring density is gram per centimeter³ (milliliter)

* Lets solve the problems

# A container of volume 150 cm³ can hold 23.6 g of aluminum

∵ Density = mass/volume

∵ The mass = 23.6 g

∵ The volume = 150 cm³

∴ The density = 23.6/150 = 59/375 ≅ 0.157 g/cm³

* The density of the aluminum is 59/375 ≅ 0.157 g/cm³

# The density of mercury is 13.6 g/cm³ at 23 C°

- A container can hold 3.2 g of mercury at this temperature

∵ Density = mass/volume

∵ The mass = 3.2 g

∵ The density = 13.6 g/cm³

∴ 13.6 = 3.2/volume

- By using cross multiplication

∴ The volume = 3.2/13.6 = 4/17 cm³

* The volume of the container is 4/17 ≅ 0.235 cm³

# The density of the iron is 4.9 g/cm³

- The beaker has a volume 250 ml

∵ Density = mass/volume

∵ The ml = cm³

∴ The volume of the beaker is 250 cm³

∵ The density = 4.9 g/cm³

∴ 4.9 = mass/250

- Multiply both sides by 250

∴ mass = 1225 g

* The mass of iron is 1225 g

7 0
3 years ago
Please help I am doing this in class right now
MAVERICK [17]

Answer:

I believe the answer is (9, -13.5)

Step-by-step explanation:

you do 6×1.5 and -9×1.5 to enlarge/dialate the quadrilateral

again I believe this is right so sorry if I am wrong

3 0
3 years ago
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