53:26 Which represents a quadratic function? f(x) = 2x3 + 2x2 – 4 f(x) = –7x2 – x + 2 f(x) = –3x + 2 f(x) = 0x2 + 3x – 3

f(x)=-7x2-x+2 represents a quadratic function.

Step-by-step explanation:

Lets do process of elimination, but first we need to know some stuff about a quadratic function.

The Standard form of a quadratic function is ax2 +bx+c where a cannot be 0.

The reason why A cannot be 0 is that if it were 0, it would make the function linear. You would just be left with bx+c, which is linear.

Alright, now lets eliminate some answers.

A f(x)=2x3+2x2-4. This is not correct.

With a quadratic function, the highest exponent needs to be 2, and it is always 2.

With this one, there is a power of 3, so this is wrong. This is a cubic function because it is raised to the power of 3.

C. f(x)=-3x+2.

This is a linear function because it is written in the y=mx+b, and it has a constant rate of change. When graphed, this would not be a parabola, or the graph of a quadratic function. Therefore, this is incorrect.

D.  f(x)=0x2+3x-3.

This is incorrect. Yes, it is true that a quadratic function in standard form is ax2+bx+c, however in this case, a=0. We know that a cannot be 0 in a quadratic function because it would make the graph linear or form a straight line.

B. f(x)= -7x2-x+2.

This answer is correct because with have an a value that is not 0, and when graphed, it forms a parabola, which is the graph of a quadratic function. In addition, the highest power is also 2.

Hope this helps, and I hope you have a good day!

∠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.

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation: