Since the vertical coordinates of each point are given by y, solve each equation for y and plug in 6 for x, as follows:

Taking the difference of the resulting y -values give the vertical distance between the points (6,27) and (6,48), which is 21.

To test the coordinates, plug the x-coordinate into the line equation and solve for y.

y = 1/3x -7

Test (3,-6)

y = 1/3(3) – 7 = 1 – 7 = -6   YES!

Test (3,7)

y = 1/3(3) – 7 = 1 – 7 = -6  NO

Test (6,-12)

y = 1/3(6) – 7 = 2 – 7 = -5  NO

Test (6,5)

y = 1/3(6) – 7 = 2 – 7 = -5  NO

Test (9,5)

y = 1/3(9) – 7 = 3 – 7 = -4  NO

Since we have –2x and +2x in the equations, it makes sense to add the equations together to give 8y = 16 yielding y = 2.  Then we substitute y = 2 into one of the original equations to get x = –2.  So the solution to the system of equations is (–2, 2)

 when plugged in for  and  make the linear equation true, therefore those coordinates fall on that line.

Because this equation is true, the point must lie on the line. The other given answer choices do not result in true equalities.

We are looking for an ordered pair that makes the given equation true. To solve, plug in the various answer choices to find the true equality.

Because this equality is true, we can conclude that the point lies on this line. None of the other given answer options will result in a true equality.

To figure out if any of the points are on the line, substitute the  and  coordinates into the equation. If the equation is incorrect, the point is not on the line. For the point :

So,  is not on the line.

If two lines intersect, that means that at one point, the  and  values are the same. Therefore, we can use substitution to solve this problem. 

Let's substitute  in for  in the other equation. Then, solve for :

Now, we can substitute this into either equation and solve for :

With these two values, the point of intersection is 

If two lines intersect, that means that their  and  values are the same at one point. Therefore, we can use substitution to solve this problem.

First, let's write these two formulas in slope-intercept form. First:

Then, for the second line:

Now, we can substitute  in for  in our second equation and solve for , like so:

Now, we can substitute this value into either equation to solve for . 

Therefore, our point of intersection is 

Since these are complementary angles, we can set up the following equation.

Now we will use the quadratic formula to solve for .

This question is testing one's ability to compare the properties of functions when they are illustrated in different forms. This question specifically is asking for the examination and interpretation of two quadratic functions for which one is illustrated in a table format and the other is illustrated graphically.

Step 1: Identify the minimum of the table.

Using the table find the time value where the lowest distance exists. 

Recall that the time represents the  values while the distance represents the  values. Therefore the ordered pair for the minimum can be written as .

Step 2: Identify the minimum of the graph

Recall that the minimum of a cubic function is known as a local minimum. This occurs at the valley where the vertex lies.

For this particular graph the vertex is at .

Step 3: Compare the minimums from step 1 and step 2.

Compare the  value coordinate from both minimums.

Therefore, the graph has the lowest minimum.