Answer :
To identify the slope of a linear relationship, we need to determine the consistent rate of change between the [tex]\( x \)[/tex]-values and the [tex]\( y \)[/tex]-values. This means we need to calculate the difference in the [tex]\( y \)[/tex]-values ([tex]\(\Delta y\)[/tex]) and the difference in the [tex]\( x \)[/tex]-values ([tex]\(\Delta x\)[/tex]), and then find the ratio [tex]\(\frac{\Delta y}{\Delta x}\)[/tex].
Here's a step-by-step approach:
1. List the given pairs of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & 7 \\ \hline 4 & 5 \\ \hline 0 & 1 \\ \hline -2 & -1 \\ \hline \end{array} \][/tex]
2. Calculate [tex]\(\Delta y\)[/tex] (difference in [tex]\( y \)[/tex]-values):
[tex]\[ \Delta y_1 = y_2 - y_1 = 5 - 7 = -2 \][/tex]
[tex]\[ \Delta y_2 = y_3 - y_2 = 1 - 5 = -4 \][/tex]
[tex]\[ \Delta y_3 = y_4 - y_3 = -1 - 1 = -2 \][/tex]
So, [tex]\(\Delta y = [-2, -4, -2]\)[/tex].
3. Calculate [tex]\(\Delta x\)[/tex] (difference in [tex]\( x \)[/tex]-values):
[tex]\[ \Delta x_1 = x_2 - x_1 = 4 - 6 = -2 \][/tex]
[tex]\[ \Delta x_2 = x_3 - x_2 = 0 - 4 = -4 \][/tex]
[tex]\[ \Delta x_3 = x_4 - x_3 = -2 - 0 = -2 \][/tex]
So, [tex]\(\Delta x = [-2, -4, -2]\)[/tex].
4. Calculate the slope (ratio [tex]\(\frac{\Delta y}{\Delta x}\)[/tex]):
[tex]\[ \text{slope}_1 = \frac{\Delta y_1}{\Delta x_1} = \frac{-2}{-2} = 1.0 \][/tex]
[tex]\[ \text{slope}_2 = \frac{\Delta y_2}{\Delta x_2} = \frac{-4}{-4} = 1.0 \][/tex]
[tex]\[ \text{slope}_3 = \frac{\Delta y_3}{\Delta x_3} = \frac{-2}{-2} = 1.0 \][/tex]
The slopes are consistent and all equal to [tex]\( 1.0 \)[/tex].
5. Conclusion:
Since all calculated slopes are the same, the common slope of the linear relationship is:
[tex]\[ \boxed{1} \][/tex]
Here's a step-by-step approach:
1. List the given pairs of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & 7 \\ \hline 4 & 5 \\ \hline 0 & 1 \\ \hline -2 & -1 \\ \hline \end{array} \][/tex]
2. Calculate [tex]\(\Delta y\)[/tex] (difference in [tex]\( y \)[/tex]-values):
[tex]\[ \Delta y_1 = y_2 - y_1 = 5 - 7 = -2 \][/tex]
[tex]\[ \Delta y_2 = y_3 - y_2 = 1 - 5 = -4 \][/tex]
[tex]\[ \Delta y_3 = y_4 - y_3 = -1 - 1 = -2 \][/tex]
So, [tex]\(\Delta y = [-2, -4, -2]\)[/tex].
3. Calculate [tex]\(\Delta x\)[/tex] (difference in [tex]\( x \)[/tex]-values):
[tex]\[ \Delta x_1 = x_2 - x_1 = 4 - 6 = -2 \][/tex]
[tex]\[ \Delta x_2 = x_3 - x_2 = 0 - 4 = -4 \][/tex]
[tex]\[ \Delta x_3 = x_4 - x_3 = -2 - 0 = -2 \][/tex]
So, [tex]\(\Delta x = [-2, -4, -2]\)[/tex].
4. Calculate the slope (ratio [tex]\(\frac{\Delta y}{\Delta x}\)[/tex]):
[tex]\[ \text{slope}_1 = \frac{\Delta y_1}{\Delta x_1} = \frac{-2}{-2} = 1.0 \][/tex]
[tex]\[ \text{slope}_2 = \frac{\Delta y_2}{\Delta x_2} = \frac{-4}{-4} = 1.0 \][/tex]
[tex]\[ \text{slope}_3 = \frac{\Delta y_3}{\Delta x_3} = \frac{-2}{-2} = 1.0 \][/tex]
The slopes are consistent and all equal to [tex]\( 1.0 \)[/tex].
5. Conclusion:
Since all calculated slopes are the same, the common slope of the linear relationship is:
[tex]\[ \boxed{1} \][/tex]
