An image of a parabolic lens is projected onto a graph. the y-intercept of the graph is (0, 90), and the zeros are 5 and 9. which equation models the function? y = 90(x – 5)(x – 9) y = 2(x – 5)(x – 9) y = 90(x + 5)(x + 9) y = 2(x + 5)(x + 9)

Step-by-step explanation:

If the zeros are 5 and 9, then the equation will have the form:

y = a (x–5) (x–9)

We know the point (0, 90) is on the curve, so we can use this to find the coefficient a:

90 = a (0–5) (0–9)

90 = 45a

a = 2

y = 2 (x – 5) (x – 9)

∠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.

x = 3 ± √(5+y)