During batting practice, two pop flies are hit from the same location, 2 s apart. The paths are modeled by the equations h = -16t2 + 56t and h = -16t2 + 156t - 248, where t is the time that has passed since the first ball was hit. possible answers: A :Isabel is correct because the y-intercept of line B is (0, 9) and the value of y when x = 0 in parabola A is 9. B: Isabel is correct because the y-intercept of line B is (9, 0) and the value of x when y = 0 in parabola A is 9. C: Isabel is incorrect because the point of intersection between line B and parabola A that can be determined is the y-intercept of the equations, not the vertex. D: Isabel is incorrect because the point of intersection between line B and parabola A that can be determined is the x-intercept of the equations, not the vertex.

wow

Step-by-step explanation:

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

∠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.