Mathematics, 27.10.2020 22:40 stefan19367

Help due now ( right answer's only)

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Minimum, Vertex(1,-7) and sorry what do you mean by aos? i dont get that please type in the comments I'll answer that too. Hope you understand the rest

Step-by-step explanation:

The vertex of a parabola means the stationary point/turning point which is denoted by the following:

$x=\frac{-b}{2a} \\$

and for y we just simply put the value of x that we get into the equation so lets get started

The equation y=x^2-2x-6 is a quadratic equation

and if we compare it by the standard form y=ax^2+bx+c we get the following values

a=1 , b=-2 , c=-6

and now for the vertex

$x=\frac{-b}{2a}$

$x=\frac{-(-2)}{2(1)} \\x=\frac{2}{2} \\x=1$

and now for y

$y=x^2-2x-6\\y=(1)^2-2(1)-6\\y=1-2-6\\y=-7$

so the vertex is (1,-7)

and the graph is minimum not maximum because the value of a is greater than 0 if the value of a>0 then the graph is minimum and if the value of a<0 meaning a is negative the graph is maximum and here in the equation a=1 which means a>0 then our graph is minimum.

Answer from: adrianaglass12

hey we were talking in the comment section with someone else 16 min ago and that chat maxed out so he made another one and wants you to come back. Just letting you know.

Step-by-step explanation:

Answer from: Quest

the 4th choice is the correct one

step-by-step explanation:

*sigh* i seriously need on this question. i don't understand it. i would appreciate it if somebody

Answer from: Quest

∠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.

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