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To determine which relationship has a zero slope, we need to calculate the slopes between consecutive points for each table of [tex]\((x, y)\)[/tex] values.
First, let’s examine the first table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 2 \\ \hline -1 & 2 \\ \hline 1 & 2 \\ \hline 3 & 2 \\ \hline \end{array} \][/tex]
The slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is calculated as:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Calculating the slopes between consecutive points in the first table:
1. Between [tex]\((-3, 2)\)[/tex] and [tex]\((-1, 2)\)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 2}{-1 + 3} = \frac{0}{2} = 0 \][/tex]
2. Between [tex]\((-1, 2)\)[/tex] and [tex]\( (1, 2) \)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 2}{1 + 1} = \frac{0}{2} = 0 \][/tex]
3. Between [tex]\( (1, 2) \)[/tex] and [tex]\( (3, 2) \)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 2}{3 - 1} = \frac{0}{2} = 0 \][/tex]
All the slopes calculated for the first table are zero, indicating a constant relationship where the [tex]\(y\)[/tex]-value does not change as [tex]\(x\)[/tex] changes. Hence, the slope is zero.
Now, let’s examine the second table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 3 \\ \hline -1 & 1 \\ \hline 1 & -1 \\ \hline 3 & -3 \\ \hline \end{array} \][/tex]
Calculating the slopes between consecutive points in the second table:
1. Between [tex]\((-3, 3)\)[/tex] and [tex]\((-1, 1)\)[/tex]:
[tex]\[ \text{slope} = \frac{1 - 3}{-1 + 3} = \frac{-2}{2} = -1 \][/tex]
2. Between [tex]\((-1, 1)\)[/tex] and [tex]\( (1, -1) \)[/tex]:
[tex]\[ \text{slope} = \frac{-1 - 1}{1 + 1} = \frac{-2}{2} = -1 \][/tex]
3. Between [tex]\( (1, -1) \)[/tex] and [tex]\( (3, -3) \)[/tex]:
[tex]\[ \text{slope} = \frac{-3 - (-1)}{3 - 1} = \frac{-2}{2} = -1 \][/tex]
All the slopes calculated for the second table are [tex]\(-1\)[/tex], indicating a consistent negative slope but not zero.
Hence, the relationship that has a zero slope is the one represented by the first table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 2 \\ \hline -1 & 2 \\ \hline 1 & 2 \\ \hline 3 & 2 \\ \hline \end{array} \][/tex]
Thus, the relationship in the first table has a zero slope.
First, let’s examine the first table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 2 \\ \hline -1 & 2 \\ \hline 1 & 2 \\ \hline 3 & 2 \\ \hline \end{array} \][/tex]
The slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is calculated as:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Calculating the slopes between consecutive points in the first table:
1. Between [tex]\((-3, 2)\)[/tex] and [tex]\((-1, 2)\)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 2}{-1 + 3} = \frac{0}{2} = 0 \][/tex]
2. Between [tex]\((-1, 2)\)[/tex] and [tex]\( (1, 2) \)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 2}{1 + 1} = \frac{0}{2} = 0 \][/tex]
3. Between [tex]\( (1, 2) \)[/tex] and [tex]\( (3, 2) \)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 2}{3 - 1} = \frac{0}{2} = 0 \][/tex]
All the slopes calculated for the first table are zero, indicating a constant relationship where the [tex]\(y\)[/tex]-value does not change as [tex]\(x\)[/tex] changes. Hence, the slope is zero.
Now, let’s examine the second table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 3 \\ \hline -1 & 1 \\ \hline 1 & -1 \\ \hline 3 & -3 \\ \hline \end{array} \][/tex]
Calculating the slopes between consecutive points in the second table:
1. Between [tex]\((-3, 3)\)[/tex] and [tex]\((-1, 1)\)[/tex]:
[tex]\[ \text{slope} = \frac{1 - 3}{-1 + 3} = \frac{-2}{2} = -1 \][/tex]
2. Between [tex]\((-1, 1)\)[/tex] and [tex]\( (1, -1) \)[/tex]:
[tex]\[ \text{slope} = \frac{-1 - 1}{1 + 1} = \frac{-2}{2} = -1 \][/tex]
3. Between [tex]\( (1, -1) \)[/tex] and [tex]\( (3, -3) \)[/tex]:
[tex]\[ \text{slope} = \frac{-3 - (-1)}{3 - 1} = \frac{-2}{2} = -1 \][/tex]
All the slopes calculated for the second table are [tex]\(-1\)[/tex], indicating a consistent negative slope but not zero.
Hence, the relationship that has a zero slope is the one represented by the first table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 2 \\ \hline -1 & 2 \\ \hline 1 & 2 \\ \hline 3 & 2 \\ \hline \end{array} \][/tex]
Thus, the relationship in the first table has a zero slope.
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