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]
