A rectangle has sides measuring (2x + 7) units and (5x + 9) units.

Part A: What is the expression that represents the area of the rectangle? Show your work. (4 points)

Part B: What are the degree and classification of the expression obtained in Part A? (3 points)

Part C: How does Part A demonstrate the closure property for polynomials? (3 points)

1. The expression for the area of the rectangle is + 53x + 63.

2. The expression is second degree (quadratic) and it can be classified as binomial.

3. Part A demonstrates that the expression is closed under addition as the closure property for the polynomial, + 53x + 63.

Step-by-step explanation:

Area of a rectangle = length × width

Assuming in the given rectangle, the length is (5x + 9) units and width is (2x + 7) units.

1. Area = (5x + 9) × (2x + 7)

       = 5x(2x + 7) + 9(2x + 7)

       = + 35x + 18x + 63

      = + 53x + 63

The expression for the area of the rectangle is + 53x + 63.

2. The expression is second degree (quadratic) and it can be classified as binomial.

3. The closure property explains that when an arithmetical operation is performed on any two numbers in a set, the result is another number in the same set.

Part A demonstrates that the expression is closed under addition as the closure property for the polynomial, + 53x + 63.

hbfhv.hjgfskhgfdbmx n

step-by-step explanation:

9-15%

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.