Answer :
To determine in which tables [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex], we need to check if for each table, the ratio [tex]\( \frac{y}{x} \)[/tex] remains constant for all pairs [tex]\((x, y)\)[/tex]. When [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex], there exists a constant [tex]\( k \)[/tex] such that [tex]\( y = kx \)[/tex].
Let's analyze each table individually.
Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 3 & 9 \\ \hline 6 & 18 \\ \hline 7 & 21 \\ \hline \end{array} \][/tex]
- For [tex]\( (1, 3) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{3}{1} = 3 \)[/tex].
- For [tex]\( (3, 9) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{9}{3} = 3 \)[/tex].
- For [tex]\( (6, 18) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{18}{6} = 3 \)[/tex].
- For [tex]\( (7, 21) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{21}{7} = 3 \)[/tex].
Since the ratio [tex]\( \frac{y}{x} \)[/tex] is constant (3) for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 1.
Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & -2 \\ \hline -2 & -1 \\ \hline 2 & 1 \\ \hline 6 & 3 \\ \hline \end{array} \][/tex]
- For [tex]\( (-4, -2) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{-2}{-4} = \frac{1}{2} \)[/tex].
- For [tex]\( (-2, -1) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{-1}{-2} = \frac{1}{2} \)[/tex].
- For [tex]\( (2, 1) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{1}{2} \)[/tex].
- For [tex]\( (6, 3) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{3}{6} = \frac{1}{2} \)[/tex].
Since the ratio [tex]\( \frac{y}{x} \)[/tex] is constant ([tex]\(\frac{1}{2}\)[/tex]) for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 2.
Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 2 \\ \hline -1 & 1 \\ \hline 2 & -2 \\ \hline 5 & -5 \\ \hline \end{array} \][/tex]
- For [tex]\( (-2, 2) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{2}{-2} = -1 \)[/tex].
- For [tex]\( (-1, 1) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{1}{-1} = -1 \)[/tex].
- For [tex]\( (2, -2) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{-2}{2} = -1 \)[/tex].
- For [tex]\( (5, -5) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{-5}{5} = -1 \)[/tex].
Since the ratio [tex]\( \frac{y}{x} \)[/tex] is constant ([tex]\(-1\)[/tex]) for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 3.
Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 0 \\ \hline 0 & -1 \\ \hline 1 & -2 \\ \hline 2 & -3 \\ \hline \end{array} \][/tex]
- For [tex]\( (-1, 0) \)[/tex], we cannot calculate [tex]\( \frac{y}{x} \)[/tex] because it results in [tex]\( \frac{0}{-1} = 0 \)[/tex].
- For the other points:
- [tex]\( (1, -2) \)[/tex], [tex]\( \frac{y}{x} = \frac{-2}{1} = -2 \)[/tex]
- [tex]\( (2, -3) \)[/tex], [tex]\( \frac{y}{x} = \frac{-3}{2} = -1.5 \)[/tex]
The ratios are not consistent across all pairs. Thus, [tex]\( y \)[/tex] does not vary directly with [tex]\( x \)[/tex] in Table 4.
The tables in which [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] are:
- Table 1
- Table 2
- Table 3
Thus, the answer is:
[tex]\[ \text{Tables } 1, 2, \text{ and } 3 \][/tex]
Let's analyze each table individually.
Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 3 & 9 \\ \hline 6 & 18 \\ \hline 7 & 21 \\ \hline \end{array} \][/tex]
- For [tex]\( (1, 3) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{3}{1} = 3 \)[/tex].
- For [tex]\( (3, 9) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{9}{3} = 3 \)[/tex].
- For [tex]\( (6, 18) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{18}{6} = 3 \)[/tex].
- For [tex]\( (7, 21) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{21}{7} = 3 \)[/tex].
Since the ratio [tex]\( \frac{y}{x} \)[/tex] is constant (3) for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 1.
Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & -2 \\ \hline -2 & -1 \\ \hline 2 & 1 \\ \hline 6 & 3 \\ \hline \end{array} \][/tex]
- For [tex]\( (-4, -2) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{-2}{-4} = \frac{1}{2} \)[/tex].
- For [tex]\( (-2, -1) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{-1}{-2} = \frac{1}{2} \)[/tex].
- For [tex]\( (2, 1) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{1}{2} \)[/tex].
- For [tex]\( (6, 3) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{3}{6} = \frac{1}{2} \)[/tex].
Since the ratio [tex]\( \frac{y}{x} \)[/tex] is constant ([tex]\(\frac{1}{2}\)[/tex]) for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 2.
Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 2 \\ \hline -1 & 1 \\ \hline 2 & -2 \\ \hline 5 & -5 \\ \hline \end{array} \][/tex]
- For [tex]\( (-2, 2) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{2}{-2} = -1 \)[/tex].
- For [tex]\( (-1, 1) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{1}{-1} = -1 \)[/tex].
- For [tex]\( (2, -2) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{-2}{2} = -1 \)[/tex].
- For [tex]\( (5, -5) \)[/tex], we calculate [tex]\( \frac{y}{x} = \frac{-5}{5} = -1 \)[/tex].
Since the ratio [tex]\( \frac{y}{x} \)[/tex] is constant ([tex]\(-1\)[/tex]) for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 3.
Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 0 \\ \hline 0 & -1 \\ \hline 1 & -2 \\ \hline 2 & -3 \\ \hline \end{array} \][/tex]
- For [tex]\( (-1, 0) \)[/tex], we cannot calculate [tex]\( \frac{y}{x} \)[/tex] because it results in [tex]\( \frac{0}{-1} = 0 \)[/tex].
- For the other points:
- [tex]\( (1, -2) \)[/tex], [tex]\( \frac{y}{x} = \frac{-2}{1} = -2 \)[/tex]
- [tex]\( (2, -3) \)[/tex], [tex]\( \frac{y}{x} = \frac{-3}{2} = -1.5 \)[/tex]
The ratios are not consistent across all pairs. Thus, [tex]\( y \)[/tex] does not vary directly with [tex]\( x \)[/tex] in Table 4.
The tables in which [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] are:
- Table 1
- Table 2
- Table 3
Thus, the answer is:
[tex]\[ \text{Tables } 1, 2, \text{ and } 3 \][/tex]
