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

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

Yes, this does form a direct variation because
[tex]$\square$[/tex]



Answer :

To determine if the relationship between [tex]\( x \)[/tex] (the number of items) and [tex]\( y \)[/tex] (the amount paid) forms a direct variation, we need to follow these steps:

1. Understand Direct Variation: A relationship is a direct variation if it can be described by an equation of the form [tex]\( y = kx \)[/tex], where [tex]\( k \)[/tex] is a constant of proportionality.

2. Calculate the Constant of Proportionality: For each pair of values [tex]\( (x, y) \)[/tex], divide [tex]\( y \)[/tex] by [tex]\( x \)[/tex] to find [tex]\( k \)[/tex].

Let's go through the table values to find [tex]\( k \)[/tex]:

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

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

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

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

3. Check for Consistency: Verify if the calculated [tex]\( k \)[/tex] is the same for all pairs of [tex]\( (x, y) \)[/tex]. From the calculations above, we can see that in each case, [tex]\( k \)[/tex] is consistently [tex]\( 0.50 \)[/tex].

Since the constant of proportionality [tex]\( k \)[/tex] is the same for all pairs of [tex]\( (x, y) \)[/tex], this relationship does indeed form a direct variation.

Therefore, the relationship forms a direct variation because:
[tex]\[ \boxed{y = 0.50x} \][/tex]