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Sagot :
To determine if the table represents a linear function, we need to examine the relationship between the [tex]\( x \)[/tex]-values and the [tex]\( y \)[/tex]-values. Specifically, we should check if the differences between consecutive [tex]\( y \)[/tex]-values are consistent. If they are, the function is linear.
Here is the given table for reference:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 10 \\ \hline 3 & 15 \\ \hline 4 & 20 \\ \hline 5 & 25 \\ \hline \end{array} \][/tex]
Now, let's calculate the differences between consecutive [tex]\( y \)[/tex]-values:
- The difference between [tex]\( y_2 \)[/tex] and [tex]\( y_1 \)[/tex] is [tex]\( 10 - 5 = 5 \)[/tex].
- The difference between [tex]\( y_3 \)[/tex] and [tex]\( y_2 \)[/tex] is [tex]\( 15 - 10 = 5 \)[/tex].
- The difference between [tex]\( y_4 \)[/tex] and [tex]\( y_3 \)[/tex] is [tex]\( 20 - 15 = 5 \)[/tex].
- The difference between [tex]\( y_5 \)[/tex] and [tex]\( y_4 \)[/tex] is [tex]\( 25 - 20 = 5 \)[/tex].
All these differences are [tex]\( 5 \)[/tex], which is consistent.
Since the differences between consecutive [tex]\( y \)[/tex]-values are the same (each equal to 5), we can conclude that the function described by this table is linear.
Moreover, a linear function can be expressed in the form [tex]\( y = mx + b \)[/tex]. To find the slope ([tex]\( m \)[/tex]), we can use any two of the points given in the table. Using the points [tex]\( (1, 5) \)[/tex] and [tex]\( (2, 10) \)[/tex]:
[tex]\[ m = \frac{\Delta y}{\Delta x} = \frac{10 - 5}{2 - 1} = \frac{5}{1} = 5 \][/tex]
Knowing the slope is [tex]\( 5 \)[/tex] and using one of the points [tex]\( (1, 5) \)[/tex], we can determine the [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]). Plugging into the linear function formula [tex]\( y = mx + b \)[/tex]:
[tex]\[ 5 = 5 \cdot 1 + b \rightarrow 5 = 5 + b \rightarrow b = 0 \][/tex]
Therefore, the linear function is [tex]\( y = 5x \)[/tex].
Thus, the given table indeed represents a linear function.
Here is the given table for reference:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 10 \\ \hline 3 & 15 \\ \hline 4 & 20 \\ \hline 5 & 25 \\ \hline \end{array} \][/tex]
Now, let's calculate the differences between consecutive [tex]\( y \)[/tex]-values:
- The difference between [tex]\( y_2 \)[/tex] and [tex]\( y_1 \)[/tex] is [tex]\( 10 - 5 = 5 \)[/tex].
- The difference between [tex]\( y_3 \)[/tex] and [tex]\( y_2 \)[/tex] is [tex]\( 15 - 10 = 5 \)[/tex].
- The difference between [tex]\( y_4 \)[/tex] and [tex]\( y_3 \)[/tex] is [tex]\( 20 - 15 = 5 \)[/tex].
- The difference between [tex]\( y_5 \)[/tex] and [tex]\( y_4 \)[/tex] is [tex]\( 25 - 20 = 5 \)[/tex].
All these differences are [tex]\( 5 \)[/tex], which is consistent.
Since the differences between consecutive [tex]\( y \)[/tex]-values are the same (each equal to 5), we can conclude that the function described by this table is linear.
Moreover, a linear function can be expressed in the form [tex]\( y = mx + b \)[/tex]. To find the slope ([tex]\( m \)[/tex]), we can use any two of the points given in the table. Using the points [tex]\( (1, 5) \)[/tex] and [tex]\( (2, 10) \)[/tex]:
[tex]\[ m = \frac{\Delta y}{\Delta x} = \frac{10 - 5}{2 - 1} = \frac{5}{1} = 5 \][/tex]
Knowing the slope is [tex]\( 5 \)[/tex] and using one of the points [tex]\( (1, 5) \)[/tex], we can determine the [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]). Plugging into the linear function formula [tex]\( y = mx + b \)[/tex]:
[tex]\[ 5 = 5 \cdot 1 + b \rightarrow 5 = 5 + b \rightarrow b = 0 \][/tex]
Therefore, the linear function is [tex]\( y = 5x \)[/tex].
Thus, the given table indeed represents a linear function.
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