Use De Morgan’s law for quantified statements and the laws of propositional logic toshow the following equivalences:.a) ¬ x(P (x) ¬Q(x)) x(¬P (x) Q(x))∀∧≡∃∨b) ¬ x(¬P (x) = Q(x)) x(¬P (x) ¬Q(x))∀⇒≡∃∧c) ¬ x(¬P (x) (Q(x) ¬R(x))) x(P (x) (¬Q(x) R(x)))

∠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.

are you an artist's daughter? cuz dang you a fine piece of art, girl.

the workers who do no drive has no reasonable reason to buy gasoline due to they don't have anything to use it for

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

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