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creativ13 [48]
2 years ago
5

Given: and bisect each other. Prove: Quadrilateral ABCD is a parallelogram. Proof: Statement Reason 1. and bisect each other. gi

ven 2. AE = EC BE = ED definition of bisection 3. m∠AEB = m∠CED 4. ΔABE ≅ ΔCDE SAS criterion 5. ∠ACD ≅ ∠CAB Corresponding angles of congruent triangles are congruent. 6. converse of Alternate Interior Angles Theorem 7. m∠BEC = m∠AED Vertical Angles Theorem 8. ΔBEC ΔDEA SAS criterion for congruence 9. DBC ≅ BDA Corresponding angles of congruent triangles are congruent. 10. converse of Alternate Interior Angles Theorem 11. Quadrilateral ABCD is a parallelogram. definition of a parallelogram 1 What is the reason for step 3 of this proof? A. Alternate Interior Angles Theorem B. Corresponding angles in congruent triangles are congruent. C. For parallel lines cut by a transversal, corresponding angles are congruent. D. Vertical Angles Theorem E.

Mathematics
2 answers:
Snezhnost [94]2 years ago
7 0

Answer:

A and D

Step-by-step explanation:

kirill115 [55]2 years ago
3 0

Answer:

To Prove: Quadrilateral ABCD is a parallelogram.

Proof: In Δ ABE and ΔCDE

   1. AE = EC and BE = ED [ Diagonals bisect each other]

   2.∠ AEB = ∠ CED [ vertically opposite angles]

Δ ABE ≅ ΔCDE----------        [SAS]

∠ ACD ≅ ∠CAB   [Corresponding angles of congruent triangles are congruent⇒This statement is untrue ∴ these are alternate interior angles not corresponding angles.]

6. The converse of alternate interior interior angle theorem states that if two parallel lines are cut by a transversal then alternate interior angles are equal.


7. In ΔBEC and ΔAED

∠BEC = ∠AED  [ Vertical Angles Theorem ]

AE = EC and BE = ED [ Diagonals bisect each other]

⇒ ΔBEC≅ ΔDEA  [ SAS criterion for congruence]

9. DBC ≅ BDA  [ Corresponding angles of congruent triangles are congruent⇒This statement is untrue ∴ these are alternate interior angles not corresponding angles.]

As pair of triangles are congruent ∵ quadrilateral ABCD is a parallelogram.

Step 3 is  m∠AEB = m∠CED

These pair of angles are vertically opposite angles of ΔAEB and ΔCED.

 Option [D. Vertical Angles Theorem]  is correct.






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a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

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I think you mean 210x^6y^4

We can deduce that this term will be located somewhere in the middle. So I will calculate k= 5; k=6 \text{ and } k =7.

For k=5

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just x^{n-k} y^k

For k=6

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

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miv72 [106K]


you could use this equation to help you solve it;

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the first step is to combine like terms;

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hope it helped...if you have any concerns just let me know:) 

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