Líst the key features of the quadratic function graphed. A) x-intercepts: (0,0), y-intercept: (0,10), vertex (3,-1)
B) x-intercepts: none, y-intercept: (0,-10), vertex: (3,-1)
C) x-intercepts: (0.-10), y-intercept: none, vertex: (3,-1)
D) x-intercepts: none, y'intercept: (3,-1), vertex (0-10)

i can answer this question sweetie!

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.

step-by-step explanation:

the true ones are a b and d