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Sagot :
Let's go through each part of the question step by step.
### Part d
For the equation [tex]\( y = 2 \times x + 1 \)[/tex] when [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 2 \times 5 + 1 = 10 + 1 = 11 \][/tex]
So, [tex]\( y = 11 \)[/tex] when [tex]\( x = 5 \)[/tex].
### Part e
For the equation [tex]\( y = 8 - x \)[/tex] when [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 8 - 2 = 6 \][/tex]
So, [tex]\( y = 6 \)[/tex] when [tex]\( x = 2 \)[/tex].
### Completing Tables and Writing the Domain
#### Part a: [tex]\( y = 2 \times x + 1 \)[/tex]
Given the input values and part of the output:
[tex]\[ x = [0, 1, 2, 4, 5, 3] \][/tex]
[tex]\[ y = [1, 3, 5, 9, 11, 13] \][/tex]
Let's fill in the table:
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline Input [tex]$(x)$[/tex] & 0 & 1 & 2 & 3 & 4 & 5 & \\
\hline Output [tex]$(y)$[/tex] & 1 & 3 & 5 & 13 & 9 & 11 & \\
\hline
\end{tabular}
Domain of [tex]\(x\)[/tex]: [tex]\(\{0, 1, 2, 3, 4, 5\}\)[/tex].
#### Part b: [tex]\( y = 5 \times x \)[/tex]
Given the input values and part of the output:
[tex]\[ x = [-3, -2, -1, 2, 0, 1, 3] \][/tex]
[tex]\[ y = [-15, -10, 0, 5, 15] \][/tex]
Let's fill in the table:
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline Input [tex]$(x)$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline Output [tex]$(y)$[/tex] & -15 & -10 & -5 & 0 & 5 & 10 & 15 \\
\hline
\end{tabular}
Domain of [tex]\(x\)[/tex]: [tex]\(\{-3, -2, -1, 0, 1, 2, 3\}\)[/tex].
#### Part c: [tex]\( y = -2 \times x \)[/tex]
Given the input values and part of the output:
[tex]\[ x = [-5, -3, -1, 0, 4] \][/tex]
[tex]\[ y = [6, 2, 0, -8] \][/tex]
Let's fill in the table:
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline Input [tex]$(x)$[/tex] & -5 & -3 & -1 & 0 & 2 & 4 & \\
\hline Output [tex]$(y)$[/tex] & 10 & 6 & 2 & 0 & -4 & -8 & -10 \\
\hline
\end{tabular}
Domain of [tex]\(x\)[/tex]: [tex]\(\{-5, -3, -1, 0, 2, 4\}\)[/tex].
#### Part d: [tex]\( y = \frac{1}{2} \times x + 2 \)[/tex]
Given the input values and part of the output:
[tex]\[ x = [-13, -12, 0, 1, -11] \][/tex]
[tex]\[ y = [-4.0, 2.0, 2.5] \][/tex]
Let's fill in the table:
\begin{tabular}{|l|l|l|l|l|l|}
\hline Input [tex]$(x)$[/tex] & -13 & -12 & 0 & 1 & -11 \\
\hline Output [tex]$(y)$[/tex] & -4.5 & -4.0 & 2.0 & 2.5 & -3.5 \\
\hline
\end{tabular}
Domain of [tex]\(x\)[/tex]: [tex]\(\{-13, -12, -11, 0, 1\}\)[/tex].
This completes the solution, matching provided numerical calculations.
### Part d
For the equation [tex]\( y = 2 \times x + 1 \)[/tex] when [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 2 \times 5 + 1 = 10 + 1 = 11 \][/tex]
So, [tex]\( y = 11 \)[/tex] when [tex]\( x = 5 \)[/tex].
### Part e
For the equation [tex]\( y = 8 - x \)[/tex] when [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 8 - 2 = 6 \][/tex]
So, [tex]\( y = 6 \)[/tex] when [tex]\( x = 2 \)[/tex].
### Completing Tables and Writing the Domain
#### Part a: [tex]\( y = 2 \times x + 1 \)[/tex]
Given the input values and part of the output:
[tex]\[ x = [0, 1, 2, 4, 5, 3] \][/tex]
[tex]\[ y = [1, 3, 5, 9, 11, 13] \][/tex]
Let's fill in the table:
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline Input [tex]$(x)$[/tex] & 0 & 1 & 2 & 3 & 4 & 5 & \\
\hline Output [tex]$(y)$[/tex] & 1 & 3 & 5 & 13 & 9 & 11 & \\
\hline
\end{tabular}
Domain of [tex]\(x\)[/tex]: [tex]\(\{0, 1, 2, 3, 4, 5\}\)[/tex].
#### Part b: [tex]\( y = 5 \times x \)[/tex]
Given the input values and part of the output:
[tex]\[ x = [-3, -2, -1, 2, 0, 1, 3] \][/tex]
[tex]\[ y = [-15, -10, 0, 5, 15] \][/tex]
Let's fill in the table:
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline Input [tex]$(x)$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline Output [tex]$(y)$[/tex] & -15 & -10 & -5 & 0 & 5 & 10 & 15 \\
\hline
\end{tabular}
Domain of [tex]\(x\)[/tex]: [tex]\(\{-3, -2, -1, 0, 1, 2, 3\}\)[/tex].
#### Part c: [tex]\( y = -2 \times x \)[/tex]
Given the input values and part of the output:
[tex]\[ x = [-5, -3, -1, 0, 4] \][/tex]
[tex]\[ y = [6, 2, 0, -8] \][/tex]
Let's fill in the table:
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline Input [tex]$(x)$[/tex] & -5 & -3 & -1 & 0 & 2 & 4 & \\
\hline Output [tex]$(y)$[/tex] & 10 & 6 & 2 & 0 & -4 & -8 & -10 \\
\hline
\end{tabular}
Domain of [tex]\(x\)[/tex]: [tex]\(\{-5, -3, -1, 0, 2, 4\}\)[/tex].
#### Part d: [tex]\( y = \frac{1}{2} \times x + 2 \)[/tex]
Given the input values and part of the output:
[tex]\[ x = [-13, -12, 0, 1, -11] \][/tex]
[tex]\[ y = [-4.0, 2.0, 2.5] \][/tex]
Let's fill in the table:
\begin{tabular}{|l|l|l|l|l|l|}
\hline Input [tex]$(x)$[/tex] & -13 & -12 & 0 & 1 & -11 \\
\hline Output [tex]$(y)$[/tex] & -4.5 & -4.0 & 2.0 & 2.5 & -3.5 \\
\hline
\end{tabular}
Domain of [tex]\(x\)[/tex]: [tex]\(\{-13, -12, -11, 0, 1\}\)[/tex].
This completes the solution, matching provided numerical calculations.
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