Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
To determine which tables display linear functions, we need to verify if the rate of change (or slope) between consecutive points in each table is consistent. A table represents a linear function if the slope between each pair of consecutive points is the same.
Here is a step-by-step analysis of each table:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 1.5 \\ \hline -1 & 0 \\ \hline 0 & -1.5 \\ \hline 1 & -3 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-2, 1.5)\)[/tex] and [tex]\((-1, 0)\)[/tex]:
[tex]\[ \frac{0 - 1.5}{-1 - (-2)} = \frac{-1.5}{1} = -1.5 \][/tex]
2. Slope between [tex]\((-1, 0)\)[/tex] and [tex]\((0, -1.5)\)[/tex]:
[tex]\[ \frac{-1.5 - 0}{0 - (-1)} = \frac{-1.5}{1} = -1.5 \][/tex]
3. Slope between [tex]\((0, -1.5)\)[/tex] and [tex]\((1, -3)\)[/tex]:
[tex]\[ \frac{-3 - (-1.5)}{1 - 0} = \frac{-1.5}{1} = -1.5 \][/tex]
Since the slopes are consistent, Table 1 represents a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 0 \\ \hline 0 & -2 \\ \hline 1 & -1 \\ \hline 2 & -3 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-1, 0)\)[/tex] and [tex]\((0, -2)\)[/tex]:
[tex]\[ \frac{-2 - 0}{0 - (-1)} = \frac{-2}{1} = -2 \][/tex]
2. Slope between [tex]\((0, -2)\)[/tex] and [tex]\((1, -1)\)[/tex]:
[tex]\[ \frac{-1 - (-2)}{1 - 0} = \frac{1}{1} = 1 \][/tex]
Since the slopes are not consistent, Table 2 does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 4 & -2.5 \\ \hline 5 & -5.5 \\ \hline 6 & -7.5 \\ \hline 7 & -10 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((4, -2.5)\)[/tex] and [tex]\((5, -5.5)\)[/tex]:
[tex]\[ \frac{-5.5 - (-2.5)}{5 - 4} = \frac{-3}{1} = -3 \][/tex]
2. Slope between [tex]\((5, -5.5)\)[/tex] and [tex]\((6, -7.5)\)[/tex]:
[tex]\[ \frac{-7.5 - (-5.5)}{6 - 5} = \frac{-2}{1} = -2 \][/tex]
3. Slope between [tex]\((6, -7.5)\)[/tex] and [tex]\((7, -10)\)[/tex]:
[tex]\[ \frac{-10 - (-7.5)}{7 - 6} = \frac{-2.5}{1} = -2.5 \][/tex]
Since the slopes are not consistent, Table 3 does not represent a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 6 \\ \hline -4 & 7 \\ \hline -5 & 8 \\ \hline -6 & 9 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-3, 6)\)[/tex] and [tex]\((-4, 7)\)[/tex]:
[tex]\[ \frac{7 - 6}{-4 - (-3)} = \frac{1}{-1} = -1 \][/tex]
2. Slope between [tex]\((-4, 7)\)[/tex] and [tex]\((-5, 8)\)[/tex]:
[tex]\[ \frac{8 - 7}{-5 - (-4)} = \frac{1}{-1} = -1 \][/tex]
3. Slope between [tex]\((-5, 8)\)[/tex] and [tex]\((-6, 9)\)[/tex]:
[tex]\[ \frac{9 - 8}{-6 - (-5)} = \frac{1}{-1} = -1 \][/tex]
Since the slopes are consistent, Table 4 represents a linear function.
Based on the analysis above, the tables that display linear functions are:
- Table 1
- Table 4
Thus, the final answer is:
[tex]\[ \boxed{\{1, 4\}} \][/tex]
Here is a step-by-step analysis of each table:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 1.5 \\ \hline -1 & 0 \\ \hline 0 & -1.5 \\ \hline 1 & -3 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-2, 1.5)\)[/tex] and [tex]\((-1, 0)\)[/tex]:
[tex]\[ \frac{0 - 1.5}{-1 - (-2)} = \frac{-1.5}{1} = -1.5 \][/tex]
2. Slope between [tex]\((-1, 0)\)[/tex] and [tex]\((0, -1.5)\)[/tex]:
[tex]\[ \frac{-1.5 - 0}{0 - (-1)} = \frac{-1.5}{1} = -1.5 \][/tex]
3. Slope between [tex]\((0, -1.5)\)[/tex] and [tex]\((1, -3)\)[/tex]:
[tex]\[ \frac{-3 - (-1.5)}{1 - 0} = \frac{-1.5}{1} = -1.5 \][/tex]
Since the slopes are consistent, Table 1 represents a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 0 \\ \hline 0 & -2 \\ \hline 1 & -1 \\ \hline 2 & -3 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-1, 0)\)[/tex] and [tex]\((0, -2)\)[/tex]:
[tex]\[ \frac{-2 - 0}{0 - (-1)} = \frac{-2}{1} = -2 \][/tex]
2. Slope between [tex]\((0, -2)\)[/tex] and [tex]\((1, -1)\)[/tex]:
[tex]\[ \frac{-1 - (-2)}{1 - 0} = \frac{1}{1} = 1 \][/tex]
Since the slopes are not consistent, Table 2 does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 4 & -2.5 \\ \hline 5 & -5.5 \\ \hline 6 & -7.5 \\ \hline 7 & -10 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((4, -2.5)\)[/tex] and [tex]\((5, -5.5)\)[/tex]:
[tex]\[ \frac{-5.5 - (-2.5)}{5 - 4} = \frac{-3}{1} = -3 \][/tex]
2. Slope between [tex]\((5, -5.5)\)[/tex] and [tex]\((6, -7.5)\)[/tex]:
[tex]\[ \frac{-7.5 - (-5.5)}{6 - 5} = \frac{-2}{1} = -2 \][/tex]
3. Slope between [tex]\((6, -7.5)\)[/tex] and [tex]\((7, -10)\)[/tex]:
[tex]\[ \frac{-10 - (-7.5)}{7 - 6} = \frac{-2.5}{1} = -2.5 \][/tex]
Since the slopes are not consistent, Table 3 does not represent a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 6 \\ \hline -4 & 7 \\ \hline -5 & 8 \\ \hline -6 & 9 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-3, 6)\)[/tex] and [tex]\((-4, 7)\)[/tex]:
[tex]\[ \frac{7 - 6}{-4 - (-3)} = \frac{1}{-1} = -1 \][/tex]
2. Slope between [tex]\((-4, 7)\)[/tex] and [tex]\((-5, 8)\)[/tex]:
[tex]\[ \frac{8 - 7}{-5 - (-4)} = \frac{1}{-1} = -1 \][/tex]
3. Slope between [tex]\((-5, 8)\)[/tex] and [tex]\((-6, 9)\)[/tex]:
[tex]\[ \frac{9 - 8}{-6 - (-5)} = \frac{1}{-1} = -1 \][/tex]
Since the slopes are consistent, Table 4 represents a linear function.
Based on the analysis above, the tables that display linear functions are:
- Table 1
- Table 4
Thus, the final answer is:
[tex]\[ \boxed{\{1, 4\}} \][/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
