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Sagot :
To identify which linear function represents a slope of [tex]\(\frac{1}{4}\)[/tex], we need to determine the slope of the line segments between consecutive points for each set of points and compare them to the desired slope.
The slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Let's analyze each set of points step-by-step.
### First Set of Points
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 3 & -11 \\ \hline 6 & 1 \\ \hline 9 & 13 \\ \hline 12 & 25 \\ \hline \end{array} \][/tex]
1. Between [tex]\((3, -11)\)[/tex] and [tex]\((6, 1)\)[/tex]:
[tex]\[ \text{slope} = \frac{1 - (-11)}{6 - 3} = \frac{1 + 11}{6 - 3} = \frac{12}{3} = 4 \][/tex]
2. Between [tex]\((6, 1)\)[/tex] and [tex]\((9, 13)\)[/tex]:
[tex]\[ \text{slope} = \frac{13 - 1}{9 - 6} = \frac{12}{3} = 4 \][/tex]
3. Between [tex]\((9, 13)\)[/tex] and [tex]\((12, 25)\)[/tex]:
[tex]\[ \text{slope} = \frac{25 - 13}{12 - 9} = \frac{12}{3} = 4 \][/tex]
The slopes between consecutive points in the first set are all [tex]\(4\)[/tex]. Since these slopes are not [tex]\(\frac{1}{4}\)[/tex], the first set of points does not represent a linear function with a slope of [tex]\(\frac{1}{4}\)[/tex].
### Second Set of Points
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -5 & 32 \\ \hline -1 & 24 \\ \hline 3 & 16 \\ \hline 7 & 8 \\ \hline \end{array} \][/tex]
1. Between [tex]\((-5, 32)\)[/tex] and [tex]\((-1, 24)\)[/tex]:
[tex]\[ \text{slope} = \frac{24 - 32}{-1 - (-5)} = \frac{24 - 32}{-1 + 5} = \frac{-8}{4} = -2 \][/tex]
2. Between [tex]\((-1, 24)\)[/tex] and [tex]\((3, 16)\)[/tex]:
[tex]\[ \text{slope} = \frac{16 - 24}{3 - (-1)} = \frac{16 - 24}{3 + 1} = \frac{-8}{4} = -2 \][/tex]
3. Between [tex]\((3, 16)\)[/tex] and [tex]\((7, 8)\)[/tex]:
[tex]\[ \text{slope} = \frac{8 - 16}{7 - 3} = \frac{8 - 16}{7 - 3} = \frac{-8}{4} = -2 \][/tex]
The slopes between consecutive points in the second set are all [tex]\(-2\)[/tex]. Since these slopes are not [tex]\(\frac{1}{4}\)[/tex], the second set of points does not represent a linear function with a slope of [tex]\(\frac{1}{4}\)[/tex].
### Conclusion
Neither set of points represents a linear function with a slope of [tex]\(\frac{1}{4}\)[/tex]. The slopes calculated for both sets do not match the required slope of [tex]\(\frac{1}{4}\)[/tex].
The slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Let's analyze each set of points step-by-step.
### First Set of Points
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 3 & -11 \\ \hline 6 & 1 \\ \hline 9 & 13 \\ \hline 12 & 25 \\ \hline \end{array} \][/tex]
1. Between [tex]\((3, -11)\)[/tex] and [tex]\((6, 1)\)[/tex]:
[tex]\[ \text{slope} = \frac{1 - (-11)}{6 - 3} = \frac{1 + 11}{6 - 3} = \frac{12}{3} = 4 \][/tex]
2. Between [tex]\((6, 1)\)[/tex] and [tex]\((9, 13)\)[/tex]:
[tex]\[ \text{slope} = \frac{13 - 1}{9 - 6} = \frac{12}{3} = 4 \][/tex]
3. Between [tex]\((9, 13)\)[/tex] and [tex]\((12, 25)\)[/tex]:
[tex]\[ \text{slope} = \frac{25 - 13}{12 - 9} = \frac{12}{3} = 4 \][/tex]
The slopes between consecutive points in the first set are all [tex]\(4\)[/tex]. Since these slopes are not [tex]\(\frac{1}{4}\)[/tex], the first set of points does not represent a linear function with a slope of [tex]\(\frac{1}{4}\)[/tex].
### Second Set of Points
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -5 & 32 \\ \hline -1 & 24 \\ \hline 3 & 16 \\ \hline 7 & 8 \\ \hline \end{array} \][/tex]
1. Between [tex]\((-5, 32)\)[/tex] and [tex]\((-1, 24)\)[/tex]:
[tex]\[ \text{slope} = \frac{24 - 32}{-1 - (-5)} = \frac{24 - 32}{-1 + 5} = \frac{-8}{4} = -2 \][/tex]
2. Between [tex]\((-1, 24)\)[/tex] and [tex]\((3, 16)\)[/tex]:
[tex]\[ \text{slope} = \frac{16 - 24}{3 - (-1)} = \frac{16 - 24}{3 + 1} = \frac{-8}{4} = -2 \][/tex]
3. Between [tex]\((3, 16)\)[/tex] and [tex]\((7, 8)\)[/tex]:
[tex]\[ \text{slope} = \frac{8 - 16}{7 - 3} = \frac{8 - 16}{7 - 3} = \frac{-8}{4} = -2 \][/tex]
The slopes between consecutive points in the second set are all [tex]\(-2\)[/tex]. Since these slopes are not [tex]\(\frac{1}{4}\)[/tex], the second set of points does not represent a linear function with a slope of [tex]\(\frac{1}{4}\)[/tex].
### Conclusion
Neither set of points represents a linear function with a slope of [tex]\(\frac{1}{4}\)[/tex]. The slopes calculated for both sets do not match the required slope of [tex]\(\frac{1}{4}\)[/tex].
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