The distance versus time plot for a particular object shows a quadratic relationship. which column of distance data is possible for this situation?
time (s) / a. distance (m) /b. distance (m) /c. distance (m) / d. distance (m) / e. distance(m)
0 / 0 / 2.00 / 9.00 / infinity / infinity
1 / 1.00 / 4.00 / 18.00 / 1.00 / 1.00
2 / 4.00 / 6.00 / 27.00 / 0.50 / 0.25
3 / 9.00 / 8.00 / 36.00 / 0.33 / 0.11
4 / 16.00 / 10.00 / 45.00 / 0.25 / 0.06
5 / 25.00 / 12.00 / 54.00 / 0.20 / 0.04
6 / 36.00 / 14.00 / 63.00 / 0.16 / 0.02
(a) column a
(b) column b
(c) column c
(d) column d
(e) column e

(A) column A

Step-by-step explanation:

The distance values in column A are:

0, 1, 4, 9, 16, 25 and 36

If you just look at these values, these are the squares of 0, 1, 2, 3, 4, 5 and 6 which shows that distance in column A versus time would show a quadratic plot.

For the values showing a quadratic relationship, the second differences in the values are constant. This means, if we find the differences of two consecutive terms for all the terms and then find the difference of the answers resulted in previous step, these should be a constant for the quadratic function.

Differences in the values of Columns A are:

1, 3, 5, 7, 9, 11

The difference in these differences are:

2, 2, 2, 2, 2

which is a constant. Since, the difference is being calculated two times, it is known as second difference. The second difference must be constant for a quadratic relationship. For the other columns, the second differences are not constant.

Therefore, the correct answer is option A.

positive

step-by-step explanation:

the quadrant is facing upwards

can you put the picture up?

step-by-step explanation: