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r-ruslan [8.4K]
3 years ago
6

1350 to the nearest 100

Mathematics
1 answer:
nydimaria [60]3 years ago
3 0

Answer:

1350 to the nearest 100 is 1400

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Which piece of additional information can be used to prove △CEA ~ △CDB?
Slav-nsk [51]

<u>Answer-</u>

<em>The correct answer is</em>

<em>∠BDC and ∠AED are right angles</em>

<u>Solution-</u>

In the ΔCEA and ΔCDB,

m\angle BCD=m\angle ACE

As this common to both of the triangle.

If ∠BDC and ∠AED are right angles, then m\angle E=90=m\angle D

Now as

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∠DBC and ∠EAC will be same. (as sum of 3 angles in a triangle is 180°)

Then, ΔCEA ≈ ΔCDB

Therefore, additional information can be used to prove ΔCEA ≈ ΔCDB is ∠BDC and ∠AED are right angles.


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garri49 [273]

Answer:

tetrahedral

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According to the valence shell electron pair repulsion theory (VSEPR) the shape of a molecule is dependent on the number of electron pairs on the valence shell of the central atom in the molecule.

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If we consider each nitrogen atom in N2 independently, we will notice that each nitrogen atom has four regions of electron density. Hence the electron pair geometry is tetrahedral.

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