Given
The vertex ( -2 , 5 ) and a random point ( -4, 3 )
Find the quadratic equation

quadratic equation

y=-1/32(x+2)²+5

Step-by-step explanation:

we have

let vertex be (p,q)=(-2,5)

random point (x,y)=(-4,3)

we have

y=a(x-p)²+q

3=a(6-(-2))²+5

3=64a+5

3-5=64a

a=-2/64=-1/32

now

quadratic equation

y=-1/32(x+2)²+5

∠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.

21

step-by-step explanation: