For this case we have the following equation:

From here, we must substitute ordered pairs of the form:
(x, y)
If the ordered pair satisfies the equation, then it belongs to the line.
We have then:
For (8, 5):
We substitute the following values:

We observe that the equation is not satisfied and therefore, this point does not belong to the line.
Since one of the points does not belong to the line, then the equation is not a good model.
Answer:
It is not a good model. One of the points does not belong to the line.
Answer: ∠ACB ≅ ∠E'C'D'; translate point D' to point B
Step-by-step explanation:
That is just my best guess.
Answer:
The answer is H 3
Step-by-step explanation:
Option B
Trevor isn't correct because -2i must also be a root
<em><u>Solution:</u></em>
For the polynomial with roots -7, 2i and 7 their roots can be,
1. ) Real roots
2.) Imaginary roots
The real roots are: -7 and +7
The imaginary root given is: 2i
The imaginary roots come from the square root. So they will be in form of 
Therefore,
For f(x) with roots -7 and +7 and
we have,

Fundamental Theorem of Algebra states that a polynomial will have exactly as many roots as its degree (the degree is the highest exponent of the polynomial).
So for f(x) with 4 roots, degree of f(x) is 4
So option B is correct. Trevor is not correct because –2i must also be a root.
The polynomial <span>3x2y2 − 5xy2 − 3x2y2 + 2x2 can be simplified by combining like terms.
The result is:
-5xy2 + 2x2
The polynomial is
a binomial (2 terms),
the degrees is 3
the highest order in x is 2 and the highest order in y is 2.</span>