Part A: ANALYZE Cameron said that for any relation, the number of elements in the domain must be greater than or equal to the number of elements in the range. Do you agree? Part B: If so, explain why. If not, give a counterexample.

In the relation {(, ), (, )}, the domain is {1}, so it has one element, while the range is {2, 3}, which has two elements.

Part A:  

No, in a function, the number of elements in the range can not be greater than the number of elements in the domain. If this were true, that would mean that an element in the domain would be associated with more than one element in the range, in which case it could not be considered a function.

Part B



Step-by-step explanation:

∠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.

what statement? is there an image?