Answer :
To determine which relationship has a zero slope, we need to calculate the slope for each of the sets of points given in the two tables.
### Relationship 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 2 \\ \hline -1 & 2 \\ \hline 1 & 2 \\ \hline 3 & 2 \\ \hline \end{array} \][/tex]
We can calculate the slope between any two points using the formula for the slope \( m \):
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
#### Slope calculations:
1. Between points \((-3, 2)\) and \((-1, 2)\):
[tex]\[ m_1 = \frac{2 - 2}{-1 - (-3)} = \frac{0}{2} = 0 \][/tex]
2. Between points \((-3, 2)\) and \( (1, 2)\):
[tex]\[ m_2 = \frac{2 - 2}{1 - (-3)} = \frac{0}{4} = 0 \][/tex]
3. Between points \((-3, 2)\) and \((3, 2)\):
[tex]\[ m_3 = \frac{2 - 2}{3 - (-3)} = \frac{0}{6} = 0 \][/tex]
The slope for each pair of points in Relationship 1 is \(0\).
### Relationship 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 3 \\ \hline -1 & 1 \\ \hline 1 & -1 \\ \hline 3 & -3 \\ \hline \end{array} \][/tex]
#### Slope calculations:
1. Between points \((-3, 3)\) and \((-1, 1)\):
[tex]\[ m_1 = \frac{1 - 3}{-1 - (-3)} = \frac{-2}{2} = -1 \][/tex]
2. Between points \((-3, 3)\) and \( (1, -1)\):
[tex]\[ m_2 = \frac{-1 - 3}{1 - (-3)} = \frac{-4}{4} = -1 \][/tex]
3. Between points \((-3, 3)\) and \((3, -3)\):
[tex]\[ m_3 = \frac{-3 - 3}{3 - (-3)} = \frac{-6}{6} = -1 \][/tex]
The slope for each pair of points in Relationship 2 is \(-1\).
### Conclusion:
Since the first relationship has a consistent slope of [tex]\(0\)[/tex], the relationship with a zero slope is Relationship 1.
### Relationship 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 2 \\ \hline -1 & 2 \\ \hline 1 & 2 \\ \hline 3 & 2 \\ \hline \end{array} \][/tex]
We can calculate the slope between any two points using the formula for the slope \( m \):
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
#### Slope calculations:
1. Between points \((-3, 2)\) and \((-1, 2)\):
[tex]\[ m_1 = \frac{2 - 2}{-1 - (-3)} = \frac{0}{2} = 0 \][/tex]
2. Between points \((-3, 2)\) and \( (1, 2)\):
[tex]\[ m_2 = \frac{2 - 2}{1 - (-3)} = \frac{0}{4} = 0 \][/tex]
3. Between points \((-3, 2)\) and \((3, 2)\):
[tex]\[ m_3 = \frac{2 - 2}{3 - (-3)} = \frac{0}{6} = 0 \][/tex]
The slope for each pair of points in Relationship 1 is \(0\).
### Relationship 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 3 \\ \hline -1 & 1 \\ \hline 1 & -1 \\ \hline 3 & -3 \\ \hline \end{array} \][/tex]
#### Slope calculations:
1. Between points \((-3, 3)\) and \((-1, 1)\):
[tex]\[ m_1 = \frac{1 - 3}{-1 - (-3)} = \frac{-2}{2} = -1 \][/tex]
2. Between points \((-3, 3)\) and \( (1, -1)\):
[tex]\[ m_2 = \frac{-1 - 3}{1 - (-3)} = \frac{-4}{4} = -1 \][/tex]
3. Between points \((-3, 3)\) and \((3, -3)\):
[tex]\[ m_3 = \frac{-3 - 3}{3 - (-3)} = \frac{-6}{6} = -1 \][/tex]
The slope for each pair of points in Relationship 2 is \(-1\).
### Conclusion:
Since the first relationship has a consistent slope of [tex]\(0\)[/tex], the relationship with a zero slope is Relationship 1.