IDNLearn.com: Your one-stop destination for reliable answers to diverse questions. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.

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. [tex]$\square$[/tex]


Sagot :

To determine if the relationship between the number of items [tex]\( x \)[/tex] and the amount paid [tex]\( y \)[/tex] forms a direct variation, we need to check if the ratio [tex]\(\frac{y}{x}\)[/tex] is constant for all data points. In mathematical terms, a direct variation relationship can be expressed as [tex]\( y = kx \)[/tex], where [tex]\( k \)[/tex] is a constant.

Here are the steps to verify this:

1. Calculate the ratio [tex]\(\frac{y}{x}\)[/tex] for each data point:

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

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

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

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

2. Examine the calculated ratios to determine if they are all the same:
- The ratios are [tex]\( 0.50, 0.50, 0.50, 0.50 \)[/tex], which are indeed all equal.

3. Conclusion:
Since the ratio [tex]\(\frac{y}{x}\)[/tex] is constant for all data points, [tex]\( y\)[/tex] and [tex]\(x \)[/tex] have a direct variation relationship. The constant of proportionality [tex]\( k \)[/tex] is [tex]\( 0.50 \)[/tex].

Hence, the relationship [tex]\( y = kx \)[/tex] where [tex]\( k = 0.50 \)[/tex] is confirmed for all the given data points.

Therefore, this relationship forms a direct variation and the verification confirms this.