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2 years ago

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Given: Quadrilateral PQRS is a rectangle. Prove: PR = QS Reason Statement 1. Quadrilateral PQRS is a rectangle. given 2. Rectang

le PQRS is a parallelogram. definition of a rectangle 3.QP ≅ RS QR ≅ PS 4. m∠QPS = m∠RSP = 90° definition of a rectangle 5. Δ PQS ≅ ΔSRP SAS criterion for congruence 6. PR ≅ QS Corresponding sides of congruent triangles are congruent. 7. PR = QS Congruent line segments have equal measures. What is the reason for the third step in this proof?

2 answers:

ryzh [129]2 years ago

7 0

Answer:

The reason for the third step in this proof : Opposite angles of parallelogram are equal

Step-by-step explanation :

Given: Quadrilateral PQRS is a rectangle.

To prove: PR = QS

Proof: 1. Given that Quadrilateral PQRS is a rectangle.

2. Definition of a rectangle : Rectangle PQRS is a parallelogram

3. QP ≅ RS QR ≅ PS (Opposite angles of parallelogram are equal).

4. m∠QPS = m∠RSP = 90° (definition of a rectangle)

5. Δ PQS ≅ ΔSRP (SAS criterion for congruence)

6. PR ≅ QS (Corresponding sides of congruent triangles are congruent).

7. PR = QS (Congruent line segments have equal measures).

Therefore, The reason for the third step in this proof : Opposite angles of parallelogram are equal

Alex73 [517]2 years ago

6 0

Answer:

Step-by-step explanation:

Given: Quadrilateral PQRS is a rectangle.

To prove: PR = QS

Proof: 1. Quadrilateral PQRS is a rectangle(Given).

2. Rectangle PQRS is a parallelogram (Definition of a rectangle).

3. QP ≅ RS QR ≅ PS (Opposite angles of parallelogram are equal).

4. m∠QPS = m∠RSP = 90° (definition of a rectangle)

5. Δ PQS ≅ ΔSRP (SAS criterion for congruence)

6. PR ≅ QS (Corresponding sides of congruent triangles are congruent).

7. PR = QS (Congruent line segments have equal measures).

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Step-by-step explanation:

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For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

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Answer:

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Step-by-step explanation:

Expanding the polynomial using the formula:

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Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

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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