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Sagot :
To complete the table based on the given domain and function [tex]\( y = -\frac{2}{3} x + 7 \)[/tex], let's find the corresponding [tex]\( y \)[/tex] values for given [tex]\( x \)[/tex] values, and vice versa where needed.
1. For [tex]\( x = -6 \)[/tex]:
[tex]\[ y = -\frac{2}{3}(-6) + 7 = 4 + 7 = 11.0 \][/tex]
Therefore, the value of [tex]\( y \)[/tex] when [tex]\( x = -6 \)[/tex] is [tex]\( 11.0 \)[/tex].
2. For [tex]\( y = 5 \)[/tex]:
[tex]\[ 5 = -\frac{2}{3} x + 7 \][/tex]
[tex]\[ -\frac{2}{3} x = 5 - 7 = -2 \][/tex]
[tex]\[ x = -2 \div -\frac{2}{3} = -2 \times -\frac{3}{2} = 3.0 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] when [tex]\( y = 5 \)[/tex] is [tex]\( 3.0 \)[/tex].
3. For [tex]\( x = 15 \)[/tex]:
[tex]\[ y = -\frac{2}{3} (15) + 7 = -10 + 7 = -3.0 \][/tex]
Therefore, the value of [tex]\( y \)[/tex] when [tex]\( x = 15 \)[/tex] is [tex]\( -3.0 \)[/tex].
4. For [tex]\( y = 15 \)[/tex]:
[tex]\[ 15 = -\frac{2}{3} x + 7 \][/tex]
[tex]\[ -\frac{2}{3} x = 15 - 7 = 8 \][/tex]
[tex]\[ x = 8 \div -\frac{2}{3} = 8 \times -\frac{3}{2} = -12.0 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] when [tex]\( y = 15 \)[/tex] is [tex]\( -12.0 \)[/tex].
Now, we can complete the table:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -6 & 11.0 \\ \hline 3.0 & 5 \\ \hline 15 & -3.0 \\ \hline -12.0 & 15 \\ \hline \end{tabular} \][/tex]
1. For [tex]\( x = -6 \)[/tex]:
[tex]\[ y = -\frac{2}{3}(-6) + 7 = 4 + 7 = 11.0 \][/tex]
Therefore, the value of [tex]\( y \)[/tex] when [tex]\( x = -6 \)[/tex] is [tex]\( 11.0 \)[/tex].
2. For [tex]\( y = 5 \)[/tex]:
[tex]\[ 5 = -\frac{2}{3} x + 7 \][/tex]
[tex]\[ -\frac{2}{3} x = 5 - 7 = -2 \][/tex]
[tex]\[ x = -2 \div -\frac{2}{3} = -2 \times -\frac{3}{2} = 3.0 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] when [tex]\( y = 5 \)[/tex] is [tex]\( 3.0 \)[/tex].
3. For [tex]\( x = 15 \)[/tex]:
[tex]\[ y = -\frac{2}{3} (15) + 7 = -10 + 7 = -3.0 \][/tex]
Therefore, the value of [tex]\( y \)[/tex] when [tex]\( x = 15 \)[/tex] is [tex]\( -3.0 \)[/tex].
4. For [tex]\( y = 15 \)[/tex]:
[tex]\[ 15 = -\frac{2}{3} x + 7 \][/tex]
[tex]\[ -\frac{2}{3} x = 15 - 7 = 8 \][/tex]
[tex]\[ x = 8 \div -\frac{2}{3} = 8 \times -\frac{3}{2} = -12.0 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] when [tex]\( y = 15 \)[/tex] is [tex]\( -12.0 \)[/tex].
Now, we can complete the table:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -6 & 11.0 \\ \hline 3.0 & 5 \\ \hline 15 & -3.0 \\ \hline -12.0 & 15 \\ \hline \end{tabular} \][/tex]
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