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PilotLPTM [1.2K]
3 years ago
5

Can someone help me with these 2 questions? I'll give u the brainliest.

Mathematics
1 answer:
Lady_Fox [76]3 years ago
8 0

Hello!

In geometry, the Triangle Inequality theorem states that the sum of any two sides of the triangle must be greater than the third side, the one not added to another. This must work for all 3 combos of pairs of sides.

So we can look at your first problem, two sides are 18 and 11. The first thing you want to see is how much would have to be added to 11 so it would be greater than 18, as 11 + x > 18 is one of the pairs. If we subtract 11 from both sides of that inequality, 11 + x - 11 > 18 - 11, then you get x > 7. And since there's only one selection where it has 7 < x, or 7 < x < 29, then you know your answer is that.

Next question. We can go through the answer choices. 4 would obviously not work, as 8 + 4 = 12, and that's less than 15. 7 wouldn't either, as 8 + 7 = 15, and 15 = 15, and it has to be greater. Same case for 23, as 15 + 8 = 23, and that wouldn't work. So the only choice that ends up being correct is 10. You can check that by doing, 15 + 8 = 23, 23 > 10. 15 + 10 = 25, 25 > 8. 8 + 10 = 18, 18 > 15.

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nikitadnepr [17]

Answer:

a=4

Step-by-step explanation:

what minus 4=8? its 12. 12-4=8 so

what times 3 is 12? 4.

3*4=12

12-4=8

a=4

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Answer:

The answer is B

Step-by-step explanation:

You add up all the numbers which should be 600 and divide by the total amount numbers used.

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adoni [48]

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Given that El bisects ZCEA, which statements must be
Alexxx [7]

Question: Given that BE bisects ∠CEA, which statements must be true? Select THREE options.

(See attachment below for the figure)

m∠CEA = 90°

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Answer:

m∠CEA = 90°

∠CEF is a straight angle.

∠AEF is a right angle

Step-by-step explanation:

Line AE is perpendicular to line CF, which is a straight line. This creates two right angles, <CEA and <AEF.

Angle on a straight line = 180°. Therefore, m<CEA + m<AEF = m<CEF. Each right angle measures 90°.

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m∠CEA = 90°

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∠AEF is a right angle

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3 years ago
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