Use the interactive number line to find each sum to complete the table.

\begin{tabular}{|c|c|c|}
\hline
[tex]$a$[/tex] & [tex]$b$[/tex] & [tex]$a + b$[/tex] \\
\hline
1 & 2 & 3 \\
\hline
-1 & -2 & [tex]$R$[/tex] \\
\hline
-4 & 1 & [tex]$S$[/tex] \\
\hline
-6 & -3 & [tex]$T$[/tex] \\
\hline
\end{tabular}



Answer :

Sure, let's find the sums for each row in the table step-by-step.

### Row 1: [tex]\( a = 1 \)[/tex] and [tex]\( b = 2 \)[/tex]
The sum of [tex]\( a + b \)[/tex] is:
[tex]\[ 1 + 2 = 3 \][/tex]
This value is already provided.

### Row 2: [tex]\( a = -1 \)[/tex] and [tex]\( b = -2 \)[/tex]
We need to find the sum of [tex]\( a + b \)[/tex].
[tex]\[ -1 + -2 = -1 - 2 = -3 \][/tex]
So, [tex]\( R = -3 \)[/tex].

### Row 3: [tex]\( a = -4 \)[/tex] and [tex]\( b = 1 \)[/tex]
We need to find the sum of [tex]\( a + b \)[/tex].
[tex]\[ -4 + 1 = -3 \][/tex]
So, [tex]\( S = -3 \)[/tex].

### Row 4: [tex]\( a = -6 \)[/tex] and [tex]\( b = -3 \)[/tex]
We need to find the sum of [tex]\( a + b \)[/tex].
[tex]\[ -6 + -3 = -6 - 3 = -9 \][/tex]
So, [tex]\( T = -9 \)[/tex].

Thus, the completed table is:
\begin{tabular}{|c|c|c|}
\hline
[tex]$a$[/tex] & [tex]$b$[/tex] & [tex]$a+b$[/tex] \\
\hline
1 & 2 & 3 \\
\hline
-1 & -2 & -3 \\
\hline
-4 & 1 & -3 \\
\hline
-6 & -3 & -9 \\
\hline
\end{tabular}

Summary of results:
- For [tex]\( a = -1 \)[/tex] and [tex]\( b = -2 \)[/tex], the sum [tex]\( R = -3 \)[/tex].
- For [tex]\( a = -4 \)[/tex] and [tex]\( b = 1 \)[/tex], the sum [tex]\( S = -3 \)[/tex].
- For [tex]\( a = -6 \)[/tex] and [tex]\( b = -3 \)[/tex], the sum [tex]\( T = -9 \)[/tex].

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