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Sagot :
To determine which of the given tables represents a linear function, let's analyze the changes in the \( y \) values relative to the changes in the \( x \) values in each table.
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 9 \\ \hline 3 & 5 \\ \hline 4 & 9 \\ \hline \end{array} \][/tex]
Calculate the differences in \( y \) and \( x \):
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 9 - 5 = 4 \), \( \Delta x = 2 - 1 = 1 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = 5 - 9 = -4 \), \( \Delta x = 3 - 2 = 1 \)
- Change from \( x = 3 \) to \( x = 4 \): \( \Delta y = 9 - 5 = 4 \), \( \Delta x = 4 - 3 = 1 \)
The differences in \( y \) (\(\Delta y\)) are not consistent; hence, this table does not represent a linear function.
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 10 \\ \hline 3 & -15 \\ \hline 4 & 20 \\ \hline \end{array} \][/tex]
Calculate the differences in \( y \) and \( x \):
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 10 - (-5) = 15 \), \( \Delta x = 2 - 1 = 1 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = -15 - 10 = -25 \), \( \Delta x = 3 - 2 = 1 \)
- Change from \( x = 3 \) to \( x = 4 \): \( \Delta y = 20 - (-15) = 35 \), \( \Delta x = 4 - 3 = 1 \)
The differences in \( y \) (\(\Delta y\)) are not consistent; hence, this table does not represent a linear function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 10 \\ \hline 3 & 20 \\ \hline 4 & 40 \\ \hline \end{array} \][/tex]
Calculate the differences in \( y \) and \( x \):
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 10 - 5 = 5 \), \( \Delta x = 2 - 1 = 1 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = 20 - 10 = 10 \), \( \Delta x = 3 - 2 = 1 \)
- Change from \( x = 3 \) to \( x = 4 \): \( \Delta y = 40 - 20 = 20 \), \( \Delta x = 4 - 3 = 1 \)
The differences in \( y \) (\(\Delta y\)) are not consistent; hence, this table does not represent a linear function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 0 \\ \hline \end{array} \][/tex]
Calculate the differences in \( y \) and \( x \):
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 0 - (-5) = 5 \), \( \Delta x = 2 - 1 = 1 \)
Since there is only one difference, we can directly compare it, and the difference is consistent:
Hence, Table 4 represents a linear function.
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 9 \\ \hline 3 & 5 \\ \hline 4 & 9 \\ \hline \end{array} \][/tex]
Calculate the differences in \( y \) and \( x \):
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 9 - 5 = 4 \), \( \Delta x = 2 - 1 = 1 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = 5 - 9 = -4 \), \( \Delta x = 3 - 2 = 1 \)
- Change from \( x = 3 \) to \( x = 4 \): \( \Delta y = 9 - 5 = 4 \), \( \Delta x = 4 - 3 = 1 \)
The differences in \( y \) (\(\Delta y\)) are not consistent; hence, this table does not represent a linear function.
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 10 \\ \hline 3 & -15 \\ \hline 4 & 20 \\ \hline \end{array} \][/tex]
Calculate the differences in \( y \) and \( x \):
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 10 - (-5) = 15 \), \( \Delta x = 2 - 1 = 1 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = -15 - 10 = -25 \), \( \Delta x = 3 - 2 = 1 \)
- Change from \( x = 3 \) to \( x = 4 \): \( \Delta y = 20 - (-15) = 35 \), \( \Delta x = 4 - 3 = 1 \)
The differences in \( y \) (\(\Delta y\)) are not consistent; hence, this table does not represent a linear function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 10 \\ \hline 3 & 20 \\ \hline 4 & 40 \\ \hline \end{array} \][/tex]
Calculate the differences in \( y \) and \( x \):
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 10 - 5 = 5 \), \( \Delta x = 2 - 1 = 1 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = 20 - 10 = 10 \), \( \Delta x = 3 - 2 = 1 \)
- Change from \( x = 3 \) to \( x = 4 \): \( \Delta y = 40 - 20 = 20 \), \( \Delta x = 4 - 3 = 1 \)
The differences in \( y \) (\(\Delta y\)) are not consistent; hence, this table does not represent a linear function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 0 \\ \hline \end{array} \][/tex]
Calculate the differences in \( y \) and \( x \):
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 0 - (-5) = 5 \), \( \Delta x = 2 - 1 = 1 \)
Since there is only one difference, we can directly compare it, and the difference is consistent:
Hence, Table 4 represents a linear function.
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