Answer :
To determine which linear function has the same slope as the one represented by the points in the table, let's analyze the slopes between each pair of points provided in the table:
Given points:
[tex]\[ (x_1, y_1) = \left( -\frac{1}{2}, \frac{1}{5} \right) \][/tex]
[tex]\[ (x_2, y_2) = \left( -\frac{1}{5}, \frac{7}{50} \right) \][/tex]
[tex]\[ (x_3, y_3) = \left( \frac{1}{5}, \frac{3}{50} \right) \][/tex]
[tex]\[ (x_4, y_4) = \left( \frac{1}{2}, 0 \right) \][/tex]
The slope [tex]\( m \)[/tex] of a line through 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 calculate the slopes between each pair of consecutive points from the table:
1. Between [tex]\((-1/2, 1/5)\)[/tex] and [tex]\((-1/5, 7/50)\)[/tex]:
[tex]\[ m_1 = \frac{\frac{7}{50} - \frac{1}{5}}{-\frac{1}{5} - (-\frac{1}{2})} \][/tex]
Simplifying the numerator:
[tex]\[ \frac{7}{50} - \frac{1}{5} = \frac{7}{50} - \frac{10}{50} = \frac{7 - 10}{50} = \frac{-3}{50} \][/tex]
Simplifying the denominator:
[tex]\[ -\frac{1}{5} - (-\frac{1}{2}) = -\frac{1}{5} + \frac{1}{2} = -\frac{2}{10} + \frac{5}{10} = \frac{3}{10} \][/tex]
Therefore,
[tex]\[ m_1 = \frac{\frac{-3}{50}}{\frac{3}{10}} = \frac{-3}{50} \times \frac{10}{3} = -\frac{3 \times 10}{50 \times 3} = -\frac{10}{50} = -\frac{1}{5} \][/tex]
2. Between [tex]\((-1/5, 7/50)\)[/tex] and [tex]\((1/5, 3/50)\)[/tex]:
[tex]\[ m_2 = \frac{\frac{3}{50} - \frac{7}{50}}{\frac{1}{5} - (-\frac{1}{5})} \][/tex]
Simplifying the numerator:
[tex]\[ \frac{3}{50} - \frac{7}{50} = \frac{3 - 7}{50} = \frac{-4}{50} = -\frac{2}{25} \][/tex]
Simplifying the denominator:
[tex]\[ \frac{1}{5} + \frac{1}{5} = \frac{2}{5} \][/tex]
Therefore,
[tex]\[ m_2 = \frac{-\frac{2}{25}}{\frac{2}{5}} = \frac{-2}{25} \times \frac{5}{2} = -\frac{2 \times 5}{25 \times 2} = -\frac{10}{50} = -\frac{1}{5} \][/tex]
3. Between [tex]\((1/5, 3/50)\)[/tex] and [tex]\((1/2, 0)\)[/tex]:
[tex]\[ m_3 = \frac{0 - \frac{3}{50}}{\frac{1}{2} - \frac{1}{5}} \][/tex]
Simplifying the numerator:
[tex]\[ 0 - \frac{3}{50} = -\frac{3}{50} \][/tex]
Simplifying the denominator:
[tex]\[ \frac{1}{2} - \frac{1}{5} = \frac{5}{10} - \frac{2}{10} = \frac{3}{10} \][/tex]
Therefore,
[tex]\[ m_3 = \frac{-\frac{3}{50}}{\frac{3}{10}} = \frac{-3}{50} \times \frac{10}{3} = -\frac{3 \times 10}{50 \times 3} = -\frac{10}{50} = -\frac{1}{5} \][/tex]
Since all slopes [tex]\(m_1\)[/tex], [tex]\(m_2\)[/tex], and [tex]\(m_3\)[/tex] are consistent and equal to [tex]\(-\frac{1}{5}\)[/tex], we can confirm the slope of the linear function represented by the table is [tex]\(-\frac{1}{5}\)[/tex].
Among the given linear functions:
1. [tex]\( y = -\frac{1}{2} x + \frac{1}{10} \)[/tex] has a slope of [tex]\(-\frac{1}{2}\)[/tex], which does not match.
2. [tex]\( y = -\frac{1}{5} x + \frac{1}{10} \)[/tex] has a slope of [tex]\(-\frac{1}{5}\)[/tex], which matches.
Therefore, the linear function [tex]\( y = -\frac{1}{5} x + \frac{1}{10} \)[/tex] has the same slope as the one represented by the table.
Given points:
[tex]\[ (x_1, y_1) = \left( -\frac{1}{2}, \frac{1}{5} \right) \][/tex]
[tex]\[ (x_2, y_2) = \left( -\frac{1}{5}, \frac{7}{50} \right) \][/tex]
[tex]\[ (x_3, y_3) = \left( \frac{1}{5}, \frac{3}{50} \right) \][/tex]
[tex]\[ (x_4, y_4) = \left( \frac{1}{2}, 0 \right) \][/tex]
The slope [tex]\( m \)[/tex] of a line through 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 calculate the slopes between each pair of consecutive points from the table:
1. Between [tex]\((-1/2, 1/5)\)[/tex] and [tex]\((-1/5, 7/50)\)[/tex]:
[tex]\[ m_1 = \frac{\frac{7}{50} - \frac{1}{5}}{-\frac{1}{5} - (-\frac{1}{2})} \][/tex]
Simplifying the numerator:
[tex]\[ \frac{7}{50} - \frac{1}{5} = \frac{7}{50} - \frac{10}{50} = \frac{7 - 10}{50} = \frac{-3}{50} \][/tex]
Simplifying the denominator:
[tex]\[ -\frac{1}{5} - (-\frac{1}{2}) = -\frac{1}{5} + \frac{1}{2} = -\frac{2}{10} + \frac{5}{10} = \frac{3}{10} \][/tex]
Therefore,
[tex]\[ m_1 = \frac{\frac{-3}{50}}{\frac{3}{10}} = \frac{-3}{50} \times \frac{10}{3} = -\frac{3 \times 10}{50 \times 3} = -\frac{10}{50} = -\frac{1}{5} \][/tex]
2. Between [tex]\((-1/5, 7/50)\)[/tex] and [tex]\((1/5, 3/50)\)[/tex]:
[tex]\[ m_2 = \frac{\frac{3}{50} - \frac{7}{50}}{\frac{1}{5} - (-\frac{1}{5})} \][/tex]
Simplifying the numerator:
[tex]\[ \frac{3}{50} - \frac{7}{50} = \frac{3 - 7}{50} = \frac{-4}{50} = -\frac{2}{25} \][/tex]
Simplifying the denominator:
[tex]\[ \frac{1}{5} + \frac{1}{5} = \frac{2}{5} \][/tex]
Therefore,
[tex]\[ m_2 = \frac{-\frac{2}{25}}{\frac{2}{5}} = \frac{-2}{25} \times \frac{5}{2} = -\frac{2 \times 5}{25 \times 2} = -\frac{10}{50} = -\frac{1}{5} \][/tex]
3. Between [tex]\((1/5, 3/50)\)[/tex] and [tex]\((1/2, 0)\)[/tex]:
[tex]\[ m_3 = \frac{0 - \frac{3}{50}}{\frac{1}{2} - \frac{1}{5}} \][/tex]
Simplifying the numerator:
[tex]\[ 0 - \frac{3}{50} = -\frac{3}{50} \][/tex]
Simplifying the denominator:
[tex]\[ \frac{1}{2} - \frac{1}{5} = \frac{5}{10} - \frac{2}{10} = \frac{3}{10} \][/tex]
Therefore,
[tex]\[ m_3 = \frac{-\frac{3}{50}}{\frac{3}{10}} = \frac{-3}{50} \times \frac{10}{3} = -\frac{3 \times 10}{50 \times 3} = -\frac{10}{50} = -\frac{1}{5} \][/tex]
Since all slopes [tex]\(m_1\)[/tex], [tex]\(m_2\)[/tex], and [tex]\(m_3\)[/tex] are consistent and equal to [tex]\(-\frac{1}{5}\)[/tex], we can confirm the slope of the linear function represented by the table is [tex]\(-\frac{1}{5}\)[/tex].
Among the given linear functions:
1. [tex]\( y = -\frac{1}{2} x + \frac{1}{10} \)[/tex] has a slope of [tex]\(-\frac{1}{2}\)[/tex], which does not match.
2. [tex]\( y = -\frac{1}{5} x + \frac{1}{10} \)[/tex] has a slope of [tex]\(-\frac{1}{5}\)[/tex], which matches.
Therefore, the linear function [tex]\( y = -\frac{1}{5} x + \frac{1}{10} \)[/tex] has the same slope as the one represented by the table.
