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Sagot :
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 check whether [tex]\(y\)[/tex] is always proportional to [tex]\(x\)[/tex]. In mathematical terms, this means that [tex]\(y = kx\)[/tex] for some constant [tex]\(k\)[/tex], where [tex]\(k\)[/tex] is the same for all data points.
Step-by-Step Solution:
1. Identify the Data Points:
The table provides the following data points:
[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]
2. Calculate the Ratio [tex]\( \frac{y}{x} \)[/tex] for Each Data Point:
- For [tex]\( x = 1, y = 0.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{0.50}{1} = 0.50 \][/tex]
- For [tex]\( x = 2, y = 1.00 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{1.00}{2} = 0.50 \][/tex]
- For [tex]\( x = 3, y = 1.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{1.50}{3} = 0.50 \][/tex]
- For [tex]\( x = 5, y = 2.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{2.50}{5} = 0.50 \][/tex]
3. Check the Consistency of Ratios:
Observe that the ratio [tex]\( \frac{y}{x} \)[/tex] is consistent across all data points:
[tex]\[ \frac{y}{x} = 0.50 \][/tex]
Therefore, the ratio remains constant at 0.50.
4. Conclude the Direct Variation:
Since the ratio [tex]\( \frac{y}{x} \)[/tex] is always 0.50 for all given data points, it means [tex]\( y = 0.50x \)[/tex]. This confirms that [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] with a constant of proportionality [tex]\( k = 0.50 \)[/tex].
Verification:
Indeed, by verifying through consistent ratio [tex]\( \frac{y}{x} = 0.50 \)[/tex] for all pairs, we conclude that the relationship between the number of items and the amount paid forms a direct variation.
Hence, the relationship forms a direct variation. The answer is [tex]\(\boxed{\text{True}}\)[/tex].
Step-by-Step Solution:
1. Identify the Data Points:
The table provides the following data points:
[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]
2. Calculate the Ratio [tex]\( \frac{y}{x} \)[/tex] for Each Data Point:
- For [tex]\( x = 1, y = 0.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{0.50}{1} = 0.50 \][/tex]
- For [tex]\( x = 2, y = 1.00 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{1.00}{2} = 0.50 \][/tex]
- For [tex]\( x = 3, y = 1.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{1.50}{3} = 0.50 \][/tex]
- For [tex]\( x = 5, y = 2.50 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{2.50}{5} = 0.50 \][/tex]
3. Check the Consistency of Ratios:
Observe that the ratio [tex]\( \frac{y}{x} \)[/tex] is consistent across all data points:
[tex]\[ \frac{y}{x} = 0.50 \][/tex]
Therefore, the ratio remains constant at 0.50.
4. Conclude the Direct Variation:
Since the ratio [tex]\( \frac{y}{x} \)[/tex] is always 0.50 for all given data points, it means [tex]\( y = 0.50x \)[/tex]. This confirms that [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] with a constant of proportionality [tex]\( k = 0.50 \)[/tex].
Verification:
Indeed, by verifying through consistent ratio [tex]\( \frac{y}{x} = 0.50 \)[/tex] for all pairs, we conclude that the relationship between the number of items and the amount paid forms a direct variation.
Hence, the relationship forms a direct variation. The answer is [tex]\(\boxed{\text{True}}\)[/tex].
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