Let D = E = {−2, −1, 0, 1, 2} Write negations for each of the following statements and determine which is true, the given statement or its negation. a. ∀x ∈ D, ∃y ∈ E such that x + y = 1. b ∃x ∈ D such that ∀y ∈ E, x + y = y. c. ∀x ∈ D, ∃y ∈ E such that xy ≥ y. d. ∃x ∈ D such that ∀y ∈ E, x ≤ y.

See explanation below.

Step-by-step explanation:

We use laws of quantifiers to write the negations. The negation of ∀w∃zP(w,z) is ∃w∀z¬P(w,z). The negation of ∃w∀zP(w,z) is ∀w∃z¬P(w,z), for a predicate P depending on w and z.  

a) Negation: ∃x ∈ D, such that ∀y ∈ E, x + y ≠ 1. In this case the negation is true, hence the original statement is false. Take x=-2∈D, and note that x+y≤0 for all y ∈ E. Therefore x+y is never 1.

b) Negation: ∀x ∈ D, ∃y ∈ E such that x + y ≠ y. Here, the original statement is true. Take x=0∈D, then x+y=0+y=y for all y∈E.

c) Negation: ∃x ∈ D, such that ∀y ∈ E, xy < y. In this case, the original statement is true. Given x∈D, take y=x∈E. Hence xy=x²≥x for all x∈D (this inequality is true for integers, and we can check it case by case in D).

d) Negation: ∀x ∈ D, ∃y ∈ E such that x > y. In this case, the original statement is true, take x=-2∈D. Then -2≤y for all y∈E.

∠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.

1010: 3030

is it i think