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Complete the table based on the given domain: [tex]\(\{-12, -6, 3, 15\}\)[/tex]

Given function: [tex]\( y = -\frac{2}{3}x + 7 \)[/tex]

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
-6 & $\square$ \\
\hline
$\square$ & 5 \\
\hline
15 & $\square$ \\
\hline
$\square$ & 15 \\
\hline
\end{tabular}
\][/tex]

Sagot :

GinnyAnsweridn
To complete the table for the given linear function [tex]\( y = -\frac{2}{3}x + 7 \)[/tex] and domain [tex]\( \{-12, -6, 3, 15\} \)[/tex]:

1. When [tex]\( x = -6 \)[/tex]:
Calculate [tex]\( y \)[/tex] when [tex]\( x = -6 \)[/tex].

Result: [tex]\( y = 11.0 \)[/tex]

2. When [tex]\( y = 5 \)[/tex]:
Find [tex]\( x \)[/tex] such that [tex]\( y = 5 \)[/tex].

Result: [tex]\( x = 3.0 \)[/tex]

3. When [tex]\( x = 15 \)[/tex]:
Calculate [tex]\( y \)[/tex] when [tex]\( x = 15 \)[/tex].

Result: [tex]\( y = -3.0 \)[/tex]

4. When [tex]\( y = 15 \)[/tex]:
Find [tex]\( x \)[/tex] such that [tex]\( y = 15 \)[/tex].

Result: [tex]\( x = -12.0 \)[/tex]

So, the completed table is:

[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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