16. Graph y = 4 sec (θ- π/6)

∠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.

a, b, c, d are correct.

step-by-step explanation:

we have g(x) = 3 and are required to find which option is a translation of g(x).

now, 'translation' is a transformation which means to shift the function in any direction and the general form of translation is g(x) + k, where k is any real number.

so, according to the options we see that all are correct as they satisfy the general form of translation.

also, it can be seen from the graph below that the function g(x) = 3 is shifted from its position in all the four options.