Answer :
To determine which table has a constant of proportionality between [tex]\( y \)[/tex] and [tex]\( x \)[/tex] of [tex]\( \frac{1}{5} \)[/tex], we need to check if the ratio [tex]\( \frac{y}{x} \)[/tex] is consistent and equals [tex]\( \frac{1}{5} \)[/tex] for each pair of values in the table.
Table A:
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 2 & \frac{4}{5} \\ 7 & \frac{14}{5} \\ 12 & \frac{24}{5} \\ \hline \end{array} \][/tex]
Calculate the constant of proportionality for each pair in Table A:
1. For [tex]\( x = 2 \)[/tex] and [tex]\( y = \frac{4}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{4}{5}}{2} = \frac{4}{5} \times \frac{1}{2} = \frac{4}{10} = \frac{2}{5} \][/tex]
2. For [tex]\( x = 7 \)[/tex] and [tex]\( y = \frac{14}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{14}{5}}{7} = \frac{14}{5} \times \frac{1}{7} = \frac{14}{35} = \frac{2}{5} \][/tex]
3. For [tex]\( x = 12 \)[/tex] and [tex]\( y = \frac{24}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{24}{5}}{12} = \frac{24}{5} \times \frac{1}{12} = \frac{24}{60} = \frac{2}{5} \][/tex]
The ratio [tex]\( \frac{y}{x} \)[/tex] is [tex]\( \frac{2}{5} \)[/tex] in all cases for Table A. Therefore, this table does not have the constant of proportionality of [tex]\( \frac{1}{5} \)[/tex].
Table B:
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 8 & \frac{8}{5} \\ 9 & \frac{9}{5} \\ 10 & 2 \\ \hline \end{array} \][/tex]
Calculate the constant of proportionality for each pair in Table B:
1. For [tex]\( x = 8 \)[/tex] and [tex]\( y = \frac{8}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{8}{5}}{8} = \frac{8}{5} \times \frac{1}{8} = \frac{8}{40} = \frac{1}{5} \][/tex]
2. For [tex]\( x = 9 \)[/tex] and [tex]\( y = \frac{9}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{9}{5}}{9} = \frac{9}{5} \times \frac{1}{9} = \frac{9}{45} = \frac{1}{5} \][/tex]
3. For [tex]\( x = 10 \)[/tex] and [tex]\( y = 2 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{2}{10} = \frac{1}{5} \][/tex]
The ratio [tex]\( \frac{y}{x} \)[/tex] is [tex]\( \frac{1}{5} \)[/tex] in all cases for Table B. Therefore, this table has the constant of proportionality of [tex]\( \frac{1}{5} \)[/tex].
Table C:
[tex]\[ \begin{array}{|ll|} \hline x & y \\ \hline 3 & \frac{16}{5} \\ \hline \end{array} \][/tex]
Calculate the constant of proportionality for the given pair in Table C:
1. For [tex]\( x = 3 \)[/tex] and [tex]\( y = \frac{16}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{16}{5}}{3} = \frac{16}{5} \times \frac{1}{3} = \frac{16}{15} \][/tex]
The ratio [tex]\( \frac{y}{x} \)[/tex] is [tex]\( \frac{16}{15} \)[/tex] for Table C, which is not [tex]\( \frac{1}{5} \)[/tex].
Therefore, the table that has a constant of proportionality between [tex]\( y \)[/tex] and [tex]\( x \)[/tex] of [tex]\( \frac{1}{5} \)[/tex] is Table B.
Table A:
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 2 & \frac{4}{5} \\ 7 & \frac{14}{5} \\ 12 & \frac{24}{5} \\ \hline \end{array} \][/tex]
Calculate the constant of proportionality for each pair in Table A:
1. For [tex]\( x = 2 \)[/tex] and [tex]\( y = \frac{4}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{4}{5}}{2} = \frac{4}{5} \times \frac{1}{2} = \frac{4}{10} = \frac{2}{5} \][/tex]
2. For [tex]\( x = 7 \)[/tex] and [tex]\( y = \frac{14}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{14}{5}}{7} = \frac{14}{5} \times \frac{1}{7} = \frac{14}{35} = \frac{2}{5} \][/tex]
3. For [tex]\( x = 12 \)[/tex] and [tex]\( y = \frac{24}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{24}{5}}{12} = \frac{24}{5} \times \frac{1}{12} = \frac{24}{60} = \frac{2}{5} \][/tex]
The ratio [tex]\( \frac{y}{x} \)[/tex] is [tex]\( \frac{2}{5} \)[/tex] in all cases for Table A. Therefore, this table does not have the constant of proportionality of [tex]\( \frac{1}{5} \)[/tex].
Table B:
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 8 & \frac{8}{5} \\ 9 & \frac{9}{5} \\ 10 & 2 \\ \hline \end{array} \][/tex]
Calculate the constant of proportionality for each pair in Table B:
1. For [tex]\( x = 8 \)[/tex] and [tex]\( y = \frac{8}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{8}{5}}{8} = \frac{8}{5} \times \frac{1}{8} = \frac{8}{40} = \frac{1}{5} \][/tex]
2. For [tex]\( x = 9 \)[/tex] and [tex]\( y = \frac{9}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{9}{5}}{9} = \frac{9}{5} \times \frac{1}{9} = \frac{9}{45} = \frac{1}{5} \][/tex]
3. For [tex]\( x = 10 \)[/tex] and [tex]\( y = 2 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{2}{10} = \frac{1}{5} \][/tex]
The ratio [tex]\( \frac{y}{x} \)[/tex] is [tex]\( \frac{1}{5} \)[/tex] in all cases for Table B. Therefore, this table has the constant of proportionality of [tex]\( \frac{1}{5} \)[/tex].
Table C:
[tex]\[ \begin{array}{|ll|} \hline x & y \\ \hline 3 & \frac{16}{5} \\ \hline \end{array} \][/tex]
Calculate the constant of proportionality for the given pair in Table C:
1. For [tex]\( x = 3 \)[/tex] and [tex]\( y = \frac{16}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{16}{5}}{3} = \frac{16}{5} \times \frac{1}{3} = \frac{16}{15} \][/tex]
The ratio [tex]\( \frac{y}{x} \)[/tex] is [tex]\( \frac{16}{15} \)[/tex] for Table C, which is not [tex]\( \frac{1}{5} \)[/tex].
Therefore, the table that has a constant of proportionality between [tex]\( y \)[/tex] and [tex]\( x \)[/tex] of [tex]\( \frac{1}{5} \)[/tex] is Table B.
