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: B' (5,8), C' (3,1)
Step-by-step explanation:
You can use the points A' (-1,8) and A (-2,-3) to find how many units point A was translated.
-2 + 1 = -1
-3 + 11 = 8
Then, you simply add 1 to the x-values of points B and C and 11 to the y-values of points B and C to get B' (5,8), C' (3,1).
I hope this helps!
Extrapolation is the process by which a value is estimated by using a previously known set of numbers. Interpolation is where a value is estimated between two known values. In this case, if you wanted to predict the value of the y-variable when the x-variable is 15, you would be extrapolating beyond the known data-set.
The result can be shown in multiple forms.
Inequality Form:
X > 8
Interval Notation:
,
(8,∞)
Using the cosine rule we can work out the length of the side opposite the indicated right angle which came to be 15mm
then the final answer came to be 163mm^2
