Consider the function f(x) = x2 and the function g(x) shown below. How will the graph of g(x) differ from the graph of f(x)? g(x) = f(x) + 5 = x2 + 5

A.
The graph of g(x) is the graph of f(x) shifted to the right 5 units.

B.
The graph of g(x) is the graph of f(x) shifted to the left 5 units.

C.
The graph of g(x) is the graph of f(x) shifted down 5 units.

D.
The graph of g(x) is the graph of f(x) shifted up 5 units.

∠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.

c: lkllmn.

hope it's you