To determine the value of [tex]\(a\)[/tex] in the table for a linear function, we need to ensure that the rate of change, or the slope, between any two pairs of points is consistent.
The provided points in the table are:
[tex]\[
(4, 26) \quad \text{and} \quad (5, 23)
\][/tex]
First, we calculate the rate of change (slope) between these two points:
[tex]\[
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{23 - 26}{5 - 4} = \frac{-3}{1} = -3
\][/tex]
This slope must be consistent between all pairs of points for the function to be linear. Therefore, the rate of change (slope) between [tex]\( (5, 23) \)[/tex] and [tex]\( (6, a) \)[/tex] should also be [tex]\(-3\)[/tex].
Using the slope formula with these points:
[tex]\[
-3 = \frac{a - 23}{6 - 5} = \frac{a - 23}{1}
\][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[
-3 = a - 23
\][/tex]
[tex]\[
a = 23 - 3
\][/tex]
[tex]\[
a = 20
\][/tex]
Thus, the value of [tex]\(a\)[/tex] must be [tex]\(20\)[/tex] to ensure the function remains linear.
Therefore, the correct answer is:
[tex]\[ a = 20 \text{ because the rate of change is } -3. \][/tex]
