The two numbers I think is 3&6 because this is a trick question
The question is defective, or at least is trying to lead you down the primrose path.
The function is linear, so the rate of change is the same no matter what interval (section) of it you're looking at.
The "rate of change" is just the slope of the function in the section. That's
(change in f(x) ) / (change in 'x') between the ends of the section.
In Section A:Length of the section = (1 - 0) = 1f(1) = 5f(0) = 0change in the value of the function = (5 - 0) = 5Rate of change = (change in the value of the function) / (size of the section) = 5/1 = 5
In Section B:Length of the section = (3 - 2) = 1 f(3) = 15f(2) = 10change in the value of the function = (15 - 10) = 5Rate of change = (change in the value of the function) / (size of the section) = 5/1 = 5
Part A:The average rate of change of each section is 5.
Part B:The average rate of change of Section B is equal to the average rate of change of Section A.
Explanation:The average rates of change in every section are equalbecause the function is linear, its graph is a straight line,and the rate of change is just the slope of the graph.
Answer:
h(x)= x^2+11x+30
Step-by-step explanation:
A quadratic function is in the form h(x) = ax^2 + bx + c.
Since the zeros are -6 and -5, take the opposite signs and add them to the variable x separately.
It should look like this: h(x)= (x+6)(x+5)
Since this is the factored form, we have to solve this equation further.
h(x)= (x+6)(x+5)
h(x)= x^2+6x+5x+30
h(x)= x^2+11x+30
Answer: 77
Step-by-step explanation:
Just finished doing the math and the correct answer should be B: 5.3 Hope this helps:)
