The answer is a = 0, b = -5, and c= 39.
<u>Step-by-step explanation</u>:
<u>step 1</u> :
A quadratic equation means that it should have at least one squared term.
<u>step 2</u> :
The standard form is ax² + bx + c = 0.
<u>step 3</u> :
The solution to the quadratic equation is usually written in the form
x = (-b ± √(b2 − 4ac))/(2a)
where a = coefficient of x^2
b = coefficient of x
c = constant term
<u>step 4</u> :
The given equation is -5x +32.
∴ The answer is a = coefficient of x^2 = 0
b = coefficient of x = -5
c = constant term = 39
Question 1
The given system of equations is:

Equate the two equations:

Rewrite in standard form:




When we put x=0, in y=3x +6, we get:

One solution is (0,6)
When we put x=-2, into y=3x+6, we get:

Another solution is (-2,0)
The solutions are; (0,6) and (-2,0)
Question 2:
The function is

Let us put x=-x,

This gives:

We can observe that:

This is the property of an even function.
Question 3:
The given function is

The average rate of change of f(x) from x=a to x=b is given as:

This is the slope of the secant line connecting the two points on f(x)
From x=2 to x=6, the average rate of change

The average rate of change is 11
<span>We add all these numbers up.
137.50 +162.50+92.50= 392.50$
Use a cacl or by hand</span>
<h2>
The "option d:
+ 13x + 12" is a trinomial with a constant term.</h2>
Step-by-step explanation:
To check options:
a: x + 4y
Here, the coefficient of x = 1 and the coefficient of y = 4
b: 
Here, the coefficient of
= 1
c:
+ 3
+ 2y
Here, the coefficient of
= 1, the coefficient of
= 4 and the coefficient of y = 2
d:
+ 13x + 12
Here, the coefficient of
= 1, the coefficient of x = 13 and
constant term = 12
Thus, the "option d)
+ 13x + 12" is a trinomial with a constant term.
The answer is C just trust me