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Use the tables below to find [tex]\((p+q)(2)\)[/tex].

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $p(x)$ \\
\hline
4 & -1 \\
\hline
2 & 3 \\
\hline
-3 & 2 \\
\hline
\end{tabular}
\][/tex]
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $q(x)$ \\
\hline
4 & 1 \\
\hline
2 & -2 \\
\hline
-3 & 5 \\
\hline
\end{tabular}
\][/tex]

[tex]\((p+q)(2) = \square\)[/tex]

Sagot :

GinnyAnsweridn
To find [tex]\((p+q)(2)\)[/tex], we first need to look at the values of [tex]\( p(x) \)[/tex] and [tex]\( q(x) \)[/tex] at [tex]\( x = 2 \)[/tex].

From the table for [tex]\( p(x) \)[/tex]:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $p(x)$ \\ \hline 4 & -1 \\ \hline 2 & 3 \\ \hline -3 & 2 \\ \hline \end{tabular} \][/tex]
we see that [tex]\( p(2) = 3 \)[/tex].

From the table for [tex]\( q(x) \)[/tex]:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $q(x)$ \\ \hline 4 & 1 \\ \hline 2 & -2 \\ \hline -3 & 5 \\ \hline \end{tabular} \][/tex]
we see that [tex]\( q(2) = -2 \)[/tex].

Now, we find [tex]\( (p+q)(2) \)[/tex] by adding these two values:
[tex]\[ (p+q)(2) = p(2) + q(2) = 3 + (-2) = 1 \][/tex]

Therefore,
[tex]\[ (p+q)(2) = 1 \][/tex]
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