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Explain Zero Rate of Change

The table below shows the amount paid for different numbers of items.
[tex]\[
\begin{tabular}{|r|r|}
\hline
$x$ & $y$ \\
\hline
1 & 0.50 \\
\hline
2 & 1.00 \\
\hline
3 & 1.50 \\
\hline
5 & 2.50 \\
\hline
\end{tabular}
\][/tex]

Determine if this relationship forms a direct variation. Verify your answer.

[tex]$\square$[/tex]


Sagot :

To determine whether the given relationship forms a direct variation, we need to verify if the ratio [tex]\( \frac{y}{x} \)[/tex] remains constant for all pairs [tex]\((x, y)\)[/tex]. A direct variation exists if and only if [tex]\( \frac{y}{x} = k \)[/tex] for some constant [tex]\( k \)[/tex].

Given the table:

[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 0.50 \\ \hline 2 & 1.00 \\ \hline 3 & 1.50 \\ \hline 5 & 2.50 \\ \hline \end{array} \][/tex]

Let’s compute the ratio [tex]\( \frac{y}{x} \)[/tex] for each pair:

1. For [tex]\( x = 1 \)[/tex] and [tex]\( y = 0.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{0.50}{1} = 0.50 \][/tex]

2. For [tex]\( x = 2 \)[/tex] and [tex]\( y = 1.00 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{1.00}{2} = 0.50 \][/tex]

3. For [tex]\( x = 3 \)[/tex] and [tex]\( y = 1.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{1.50}{3} = 0.50 \][/tex]

4. For [tex]\( x = 5 \)[/tex] and [tex]\( y = 2.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{2.50}{5} = 0.50 \][/tex]

We observe that the ratio [tex]\( \frac{y}{x} \)[/tex] is equal to [tex]\( 0.50 \)[/tex] in each case. Because the ratio remains the same for all pairs, this indicates a direct variation relationship.

Therefore, the given relationship forms a direct variation. The constant of variation [tex]\( k \)[/tex] is [tex]\( 0.50 \)[/tex].