Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Certainly! Let's address each part of your task step-by-step. Make sure to have your graph sheet ready for visual representation and follow along with these detailed instructions.
### Part (a): Completing the Table of Values
Let's work with the equations given:
1. [tex]\( x - 2y = -2 \)[/tex]
2. [tex]\( x - 2y = 2 \)[/tex]
We need to find the [tex]\( y \)[/tex] values corresponding to [tex]\( x \)[/tex] values from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex].
#### For [tex]\( x - 2y = -2 \)[/tex]
Rearranging the equation for [tex]\( y \)[/tex]:
[tex]\[ x - 2y = -2 \][/tex]
[tex]\[ -2y = -2 - x \][/tex]
[tex]\[ y = \frac{x + 2}{2} \][/tex]
Now, let’s determine the [tex]\( y \)[/tex] values for x [tex]\( = -3, -2, -1, 0, 1, 2, 3 \)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \frac{-3 + 2}{2} = \frac{-1}{2} = -0.5 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
Given: [tex]\( y = 1 \)[/tex] already provided in the table.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = \frac{-1 + 2}{2} = \frac{1}{2} = 0.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
Given: [tex]\( y = 1 \)[/tex] already provided in the table.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \frac{1 + 2}{2} = \frac{3}{2} = 1.5 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
Given: [tex]\( y = 2 \)[/tex] already provided in the table.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{3 + 2}{2} = \frac{5}{2} = 2.5 \][/tex]
So, the completed table for [tex]\( x - 2y = -2 \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$x - 2y = -2$[/tex] & -0.5 & 1 & 0.5 & 1 & 1.5 & 2 & 2.5 \\
\hline
\end{tabular}
#### For [tex]\( x - 2y = 2 \)[/tex]
Rearranging the equation for [tex]\( y \)[/tex]:
[tex]\[ x - 2y = 2 \][/tex]
[tex]\[ -2y = 2 - x \][/tex]
[tex]\[ y = \frac{x - 2}{2} \][/tex]
Now, let’s determine the [tex]\( y \)[/tex] values for [tex]\(x\)[/tex] from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \frac{-3 - 2}{2} = \frac{-5}{2} = -2.5 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
Given: [tex]\( y = -2 \)[/tex] already provided in the table.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = \frac{-1 - 2}{2} = \frac{-3}{2} = -1.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
Given: [tex]\( y = -1 \)[/tex] already provided in the table.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \frac{1 - 2}{2} = \frac{-1}{2} = -0.5 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
Given: [tex]\( y = 0 \)[/tex] already provided in the table.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{3 - 2}{2} = \frac{1}{2} = 0.5 \][/tex]
So, the completed table for [tex]\( x - 2y = 2 \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$x - 2y = -2$[/tex] & -0.5 & 1 & 0.5 & 1 & 1.5 & 2 & 2.5 \\
[tex]$x - 2y = 2$[/tex] & -2.5 & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 \\
\hline
\end{tabular}
### Part (b): Determining the Gradient
To determine the gradient (slope) of the line [tex]\( x - 2y = 2 \)[/tex]:
Rewriting the equation in slope-intercept form [tex]\( y = mx + b \)[/tex] where [tex]\( m \)[/tex] is the slope:
[tex]\[ x - 2y = 2 \][/tex]
[tex]\[ -2y = -x + 2 \][/tex]
[tex]\[ y = \frac{1}{2}x - 1 \][/tex]
Thus, the gradient [tex]\( m \)[/tex] of the line [tex]\( x - 2y = 2 \)[/tex] is:
[tex]\[ m = \frac{1}{2} \][/tex]
### Part (a): Completing the Table of Values
Let's work with the equations given:
1. [tex]\( x - 2y = -2 \)[/tex]
2. [tex]\( x - 2y = 2 \)[/tex]
We need to find the [tex]\( y \)[/tex] values corresponding to [tex]\( x \)[/tex] values from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex].
#### For [tex]\( x - 2y = -2 \)[/tex]
Rearranging the equation for [tex]\( y \)[/tex]:
[tex]\[ x - 2y = -2 \][/tex]
[tex]\[ -2y = -2 - x \][/tex]
[tex]\[ y = \frac{x + 2}{2} \][/tex]
Now, let’s determine the [tex]\( y \)[/tex] values for x [tex]\( = -3, -2, -1, 0, 1, 2, 3 \)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \frac{-3 + 2}{2} = \frac{-1}{2} = -0.5 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
Given: [tex]\( y = 1 \)[/tex] already provided in the table.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = \frac{-1 + 2}{2} = \frac{1}{2} = 0.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
Given: [tex]\( y = 1 \)[/tex] already provided in the table.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \frac{1 + 2}{2} = \frac{3}{2} = 1.5 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
Given: [tex]\( y = 2 \)[/tex] already provided in the table.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{3 + 2}{2} = \frac{5}{2} = 2.5 \][/tex]
So, the completed table for [tex]\( x - 2y = -2 \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$x - 2y = -2$[/tex] & -0.5 & 1 & 0.5 & 1 & 1.5 & 2 & 2.5 \\
\hline
\end{tabular}
#### For [tex]\( x - 2y = 2 \)[/tex]
Rearranging the equation for [tex]\( y \)[/tex]:
[tex]\[ x - 2y = 2 \][/tex]
[tex]\[ -2y = 2 - x \][/tex]
[tex]\[ y = \frac{x - 2}{2} \][/tex]
Now, let’s determine the [tex]\( y \)[/tex] values for [tex]\(x\)[/tex] from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \frac{-3 - 2}{2} = \frac{-5}{2} = -2.5 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
Given: [tex]\( y = -2 \)[/tex] already provided in the table.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = \frac{-1 - 2}{2} = \frac{-3}{2} = -1.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
Given: [tex]\( y = -1 \)[/tex] already provided in the table.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \frac{1 - 2}{2} = \frac{-1}{2} = -0.5 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
Given: [tex]\( y = 0 \)[/tex] already provided in the table.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{3 - 2}{2} = \frac{1}{2} = 0.5 \][/tex]
So, the completed table for [tex]\( x - 2y = 2 \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$x - 2y = -2$[/tex] & -0.5 & 1 & 0.5 & 1 & 1.5 & 2 & 2.5 \\
[tex]$x - 2y = 2$[/tex] & -2.5 & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 \\
\hline
\end{tabular}
### Part (b): Determining the Gradient
To determine the gradient (slope) of the line [tex]\( x - 2y = 2 \)[/tex]:
Rewriting the equation in slope-intercept form [tex]\( y = mx + b \)[/tex] where [tex]\( m \)[/tex] is the slope:
[tex]\[ x - 2y = 2 \][/tex]
[tex]\[ -2y = -x + 2 \][/tex]
[tex]\[ y = \frac{1}{2}x - 1 \][/tex]
Thus, the gradient [tex]\( m \)[/tex] of the line [tex]\( x - 2y = 2 \)[/tex] is:
[tex]\[ m = \frac{1}{2} \][/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.