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.

The lengths of the horizontal and vertical legs of the triangle correspond to the -coordinate  of the -intercept and the -coordinate  of the -intercept. The area of a right triangle is half the product of the lengths of its legs  and . The length of the vertical leg is , so, setting  and , and solving for :

Therefore, the -intercept of the line containing the hypotenuse is . The slope of the line given the coordinates of its intercepts is 

.

substituting:

.

Substituting for  and  in the slope-intercept form of the equation of a line, 

,

the line has equation

.

Substituting  for  and 6 for  and solving for , we find the -coordinate 

of the point on the line with -coordinate 6:

Use 2 points from the chart to find the equation of the line. 

Example: (–1, 3) and (1, –1)  

Using the formula for the slope, we find the slope to be –2.  Putting that into our equation for a line we get y = –2x + b.  Plug in one of the points for x and y into this equation in order to find b.  b = 1.  

The equation then will be: y = –2x + 1. 

Plug in 5 for x in order to find y.  

y = –2(5) + 1  

y = –9

The slope of a line is defined as a change in the y coordinates over a change in the x coordinates (rise over run).

To calculate the slope of a line, use the following formula: 

The slope of the line that passes these two points are simply ∆y/∆x = (3-5)/(2+3) = -2/5

The slope is negative as it starts in quadrant 2 and ends in quadrant 4. Slope is equivlent to the change in y divided by the change in x. The change in y is greater than the change in x, which implies that the slope must be less than –1, leaving –2 as the only possible solution.