\(\begin{array}{c|c} x & 0 & 2 & 4 & 6\\ \hline y & 1 & 2 & 3 & 4\end{array}\). The table is for the relation y = mx + c where m and c are constants. What i...

Answer Details

We can use the values in the table to find the values of m and c. Since y = mx + c, we can write: When x = 0, y = c, so c = 1 When x = 2, y = 2m + 1 When x = 4, y = 4m + 1 When x = 6, y = 6m + 1 To find the value of m, we can use the values for x = 2 and x = 4: 2m + 1 = 2 4m + 1 = 3 Solving these equations simultaneously gives m = 1/2. Substituting this value of m and c = 1 into y = mx + c gives: y = (1/2)x + 1 So the equation of the line described in the table is y = (1/2)x + 1. Therefore, the answer is (d) y = \(\frac{1}{2}x + 1\).