Algebra A Unit 7 Quiz 23-24 DUE 10/30/24
2 of 142 of 14 Items
Question
POSSIBLE POINTS: 11.54
The table below shows the number of jumping jacks completed after a given period of time in minutes.
Time (minutes) Jumping Jacks
1 50
2 100
3 150
4 200
Considering the jumping jacks: 50, 100, 150, 200, what is the common difference?
Now, think of this table as a set of ordered pairs. This means that the first row can be placed in an ordered pair as (1, 50). The second row can be written as (2, 100). Using this, what is the slope of the line that connects the first two points?
What is the slope of the line that connects the 3rd and 4th point?
What is the slope of the line that connects the 1st and the 4th point?
Is the common difference (aka slope aka rate of change) constant?
Why is it or is it not constant?
The common difference can be found by subtracting the previous term from the current term.
For the jumping jacks: 100 - 50 = 50, 150 - 100 = 50, 200 - 150 = 50
So, the common difference is 50.
Now, let's find the slopes of the lines connecting the points.
The slope between the first two points (1, 50) and (2, 100) can be found using the formula: slope = (change in y)/(change in x).
Slope = (100 - 50)/(2 - 1) = 50/1 = 50
Similarly, the slope between the third and fourth point (3, 150) and (4, 200) is:
Slope = (200 - 150)/(4 - 3) = 50/1 = 50
The slope between the first and fourth point (1, 50) and (4, 200) is:
Slope = (200 - 50)/(4 - 1) = 150/3 = 50
So, the slope remains constant for all these scenarios.
The common difference (slope) is constant because the rate of change between any two points in the table is the same. In other words, for every 1 minute increase in time, there is a consistent increase of 50 jumping jacks. This suggests a linear relationship between the time and the number of jumping jacks, where the rate of change (slope) remains constant.