(5r-4)(r^2-6r+4) what is the product

Distribute sum groups

5r(r^2 - 6r + 4) -4(r^2 - 6r + 4)

Distribute

5r^3 - 30r^2 + 20r - 4(r^2 - 6r + 4)

Distribute

5r^3 - 30r^2 + 20r - (4r^2 - 24r + 16)

Simplify brackets

5r^3 - 30r^2 + 20r - 4r^2 + 24r - 16

Collect like terms

5r^3 (-30r^2 - 4r^2) + (20 + 24r) - 16

Simplify

5r^3 - 34r^2 + 44r - 16

∠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.

good luck on your studies! : )

david edward.