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nikklg [1K]
3 years ago
14

The choices are: J and L J and M K and L K and M

Mathematics
1 answer:
almond37 [142]3 years ago
7 0
I would say the answer is K and M
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Find the value of x. If​ necessary, write your answer in the simplest radical form.
Paul [167]

Answer:

x = 4\sqrt{5}

Step-by-step explanation:

Using Pythagoras' identity in the right triangle.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

x² + 1² = 9²

x² + 1 = 81 ( subtract 1 from both sides )

x² = 80 ( take square root of both sides )

x = \sqrt{80} = \sqrt{16(5)} = \sqrt{16} × \sqrt{5} = 4\sqrt{5}

3 0
2 years ago
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-2|0.25-0.5 |+2=<br> What is this answer
Anestetic [448]

1.5

...

−2∣−0.25∣+2

−2×0.25+2

−0.5+2

8 0
3 years ago
Two gears are connected and are rotating simultaneously. The smaller gear has a radius of 4 inches, and the larger gear has a ra
pychu [463]
Draw a diagram to illustrate the problem as shown below.

When the smaller gear rotates through a revolution, it sweeps an arc length of
2π(4) = 8π inches.

Part 1
The same arc length is swept by the larger gear. The central angle of the larger gear, x, is
7x = 8π
x = (8π)/7 radians = (8π)/7 * (180/π) = 205.7°

Answer: 205.7° (nearest tenth)

Part 2
When the larger gear makes one rotation, it sweeps an arc length of 
2π(7) = 14π inches.
If the central angle for the smaller gear is y radians, then
4y = 14π
y = 3.5π radians = (3.5π)/2π revolutions = 1.75 revolutions

Answer:
The smaller gear makes 1.75 rotations

7 0
3 years ago
Evaluate 40 divided by 8 plus 3 multiplied by 4 multiplied bye 2
Natali5045456 [20]

Answer:

29

Step-by-step explanation:

40/8 + 3*4*2 =

5 + 12*2 =

5 + 24 =

29

5 0
3 years ago
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The product of 5 and the cube of x, increase by the difference of 6 and x3
lawyer [7]

Answer:

4x^{3} +6

Step-by-step explanation:

  • The product of 5 and the cube of x

5*x^3\\\\=5x^3

  • the difference of 6 and x^3

6-x^3

I understand as increase as sum so

(5x^3)+(6-x^3)\\\\5x^3+6-x^3\\\\(5-1)x^3+6\\\\4x^3+6

5 0
3 years ago
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