Select the correct answer.

In which table does [tex]$y$[/tex] vary directly with [tex]$x$[/tex]?

A.
\begin{tabular}{|c|c|}
\hline
[tex]x[/tex] & [tex]y[/tex] \\
\hline
1 & -2 \\
\hline
2 & -4 \\
\hline
3 & -16 \\
\hline
\end{tabular}

B.
\begin{tabular}{|c|c|}
\hline
[tex]x[/tex] & [tex]y[/tex] \\
\hline
1 & -5 \\
\hline
2 & 18 \\
\hline
3 & 41 \\
\hline
\end{tabular}

C.
\begin{tabular}{|c|c|}
\hline
[tex]x[/tex] & [tex]y[/tex] \\
\hline
1 & 26 \\
\hline
2 & 52 \\
\hline
3 & 78 \\
\hline
\end{tabular}

D.
\begin{tabular}{|c|c|}
\hline
[tex]x[/tex] & [tex]y[/tex] \\
\hline
1 & -7 \\
\hline
2 & -1 \\
\hline
3 & 6 \\
\hline
\end{tabular}



Answer :

To determine which table shows a direct variation of [tex]\( y \)[/tex] with [tex]\( x \)[/tex], we need to check if [tex]\( y \)[/tex] is directly proportional to [tex]\( x \)[/tex]. This means that [tex]\( y = kx \)[/tex] for some constant [tex]\( k \)[/tex], where [tex]\( k \)[/tex] is the same for all pairs of [tex]\( (x, y) \)[/tex] in the table.

Let's check each table.

Table A:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -4 \\ \hline 3 & -16 \\ \hline \end{array} \][/tex]
When [tex]\( x = 1 \)[/tex], [tex]\( y = -2 \)[/tex]. \\
When [tex]\( x = 2 \)[/tex], [tex]\( y = -4 \)[/tex]. \\
When [tex]\( x = 3 \)[/tex], [tex]\( y = -16 \)[/tex]. \\
Let's calculate [tex]\( \frac{y}{x} \)[/tex]:
[tex]\[ \frac{-2}{1} = -2 \quad (k = -2) \][/tex]
[tex]\[ \frac{-4}{2} = -2 \quad (k = -2) \][/tex]
[tex]\[ \frac{-16}{3} = -\frac{16}{3} \quad (k = -\frac{16}{3}) \][/tex]
Since [tex]\( k \)[/tex] is not constant (k changes between [tex]\(-2\)[/tex] and [tex]\(-\frac{16}{3}\)[/tex]), [tex]\( y \)[/tex] does not vary directly with [tex]\( x \)[/tex].

Table B:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 18 \\ \hline 3 & 41 \\ \hline \end{array} \][/tex]
When [tex]\( x = 1 \)[/tex], [tex]\( y = -5 \)[/tex]. \\
When [tex]\( x = 2 \)[/tex], [tex]\( y = 18 \)[/tex]. \\
When [tex]\( x = 3 \)[/tex], [tex]\( y = 41 \)[/tex]. \\
Let's calculate [tex]\( \frac{y}{x} \)[/tex]:
[tex]\[ \frac{-5}{1} = -5 \quad (k = -5) \][/tex]
[tex]\[ \frac{18}{2} = 9 \quad (k = 9) \][/tex]
[tex]\[ \frac{41}{3} \approx 13.67 \quad (k \approx 13.67) \][/tex]
Since [tex]\( k \)[/tex] is not constant (k changes between [tex]\(-5\)[/tex], [tex]\(9\)[/tex], and [tex]\(\approx 13.67\)[/tex]), [tex]\( y \)[/tex] does not vary directly with [tex]\( x \)[/tex].

Table C:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 26 \\ \hline 2 & 52 \\ \hline 3 & 78 \\ \hline \end{array} \][/tex]
When [tex]\( x = 1 \)[/tex], [tex]\( y = 26 \)[/tex]. \\
When [tex]\( x = 2 \)[/tex], [tex]\( y = 52 \)[/tex]. \\
When [tex]\( x = 3 \)[/tex], [tex]\( y = 78 \)[/tex]. \\
Let's calculate [tex]\( \frac{y}{x} \)[/tex]:
[tex]\[ \frac{26}{1} = 26 \quad (k = 26) \][/tex]
[tex]\[ \frac{52}{2} = 26 \quad (k = 26) \][/tex]
[tex]\[ \frac{78}{3} = 26 \quad (k = 26) \][/tex]
Since [tex]\( k \)[/tex] is constant ([tex]\( k = 26 \)[/tex]), [tex]\( y \)[/tex] does vary directly with [tex]\( x \)[/tex].

Table D:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -7 \\ \hline 2 & -1 \\ \hline 3 & 6 \\ \hline \end{array} \][/tex]
When [tex]\( x = 1 \)[/tex], [tex]\( y = -7 \)[/tex]. \\
When [tex]\( x = 2 \)[/tex], [tex]\( y = -1 \)[/tex]. \\
When [tex]\( x = 3 \)[/tex], [tex]\( y = 6 \)[/tex]. \\
Let's calculate [tex]\( \frac{y}{x} \)[/tex]:
[tex]\[ \frac{-7}{1} = -7 \quad (k = -7) \][/tex]
[tex]\[ \frac{-1}{2} = -0.5 \quad (k = -0.5) \][/tex]
[tex]\[ \frac{6}{3} = 2 \quad (k = 2) \][/tex]
Since [tex]\( k \)[/tex] is not constant (k changes between [tex]\(-7\)[/tex], [tex]\(-0.5\)[/tex], and [tex]\(2\)[/tex]), [tex]\( y \)[/tex] does not vary directly with [tex]\( x \)[/tex].

Therefore, the correct table where [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] is:
[tex]\[ \boxed{\text{C}} \][/tex]