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\begin{tabular}{|c|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] & [tex]$(x, y)$[/tex] \\
\hline
[tex]$10+(-5)$[/tex] & [tex]$-1+3$[/tex] & [tex]$(5, 2)$[/tex] \\
\hline
[tex]$0-(-4)$[/tex] & [tex]$-6-(-6)$[/tex] & [tex]$(4, 0)$[/tex] \\
\hline
[tex]$-2 \times(-2)$[/tex] & [tex]$-1 \times 2$[/tex] & [tex]$(4, -2)$[/tex] \\
\hline
[tex]$-30 \div(-5)$[/tex] & [tex]$-18 \div 6$[/tex] & [tex]$(6, -3)$[/tex] \\
\hline
[tex]$-2+9$[/tex] & [tex]$3+(-5)$[/tex] & [tex]$(7, -2)$[/tex] \\
\hline
[tex]$-1-(-9)$[/tex] & [tex]$5-5$[/tex] & [tex]$(8, 0)$[/tex] \\
\hline
[tex]$-7 \times(-1)$[/tex] & [tex]$-2 \times(-1)$[/tex] & [tex]$(7, 2)$[/tex] \\
\hline
[tex]$-40 \div(-8)$[/tex] & [tex]$-10 \div(-5)$[/tex] & [tex]$(5, 2)$[/tex] \\
\hline
\end{tabular}



Answer :

GinnyAnswer
Certainly! Let's solve each row step by step to fill in the remaining [tex]$(x, y)$[/tex] values in the table.

1. First row (already filled in):
- [tex]\( x = 10 + (-5) = 5 \)[/tex]
- [tex]\( y = -1 + 3 = 2 \)[/tex]
- So, [tex]\((x, y) = (5, 2)\)[/tex]

2. Second row:
- [tex]\( x = 0 - (-4) \)[/tex]
- Calculating [tex]\( 0 - (-4) \)[/tex]:
- [tex]\( x = 0 + 4 = 4 \)[/tex]
- [tex]\( y = -6 - (-6) \)[/tex]
- Calculating [tex]\( -6 - (-6) \)[/tex]:
- [tex]\( y = -6 + 6 = 0 \)[/tex]
- So, [tex]\((x, y) = (4, 0)\)[/tex]

3. Third row:
- [tex]\( x = -2 \times (-2) \)[/tex]
- Calculating [tex]\(-2 \times (-2)\)[/tex]:
- [tex]\( x = 4 \)[/tex]
- [tex]\( y = -1 \times 2 \)[/tex]
- Calculating [tex]\( -1 \times 2 \)[/tex]:
- [tex]\( y = -2 \)[/tex]
- So, [tex]\((x, y) = (4, -2)\)[/tex]

4. Fourth row:
- [tex]\( x = -30 \div (-5) \)[/tex]
- Calculating [tex]\(-30 \div (-5)\)[/tex]:
- [tex]\( x = 6.0 \)[/tex]
- [tex]\( y = -18 \div 6 \)[/tex]
- Calculating [tex]\(-18 \div 6 \)[/tex]:
- [tex]\( y = -3.0 \)[/tex]
- So, [tex]\((x, y) = (6.0, -3.0)\)[/tex]

5. Fifth row:
- [tex]\( x = -2 + 9 \)[/tex]
- Calculating [tex]\( -2 + 9 \)[/tex]:
- [tex]\( x = 7 \)[/tex]
- [tex]\( y = 3 + (-5) \)[/tex]
- Calculating [tex]\( 3 + (-5) \)[/tex]:
- [tex]\( y = -2 \)[/tex]
- So, [tex]\((x, y) = (7, -2)\)[/tex]

6. Sixth row:
- [tex]\( x = -1 - (-9) \)[/tex]
- Calculating [tex]\( -1 - (-9) \)[/tex]:
- [tex]\( x = -1 + 9 = 8 \)[/tex]
- [tex]\( y = 5 - 5 \)[/tex]
- Calculating [tex]\( 5 - 5 \)[/tex]:
- [tex]\( y = 0 \)[/tex]
- So, [tex]\((x, y) = (8, 0)\)[/tex]

7. Seventh row:
- [tex]\( x = -7 \times (-1) \)[/tex]
- Calculating [tex]\(-7 \times (-1)\)[/tex]:
- [tex]\( x = 7 \)[/tex]
- [tex]\( y = -2 \times (-1) \)[/tex]
- Calculating [tex]\(-2 \times (-1)\)[/tex]:
- [tex]\( y = 2 \)[/tex]
- So, [tex]\((x, y) = (7, 2)\)[/tex]

8. Eighth row:
- [tex]\( x = -40 \div (-8) \)[/tex]
- Calculating [tex]\(-40 \div (-8)\)[/tex]:
- [tex]\( x = 5.0 \)[/tex]
- [tex]\( y = -10 \div (-5) \)[/tex]
- Calculating [tex]\(-10 \div (-5)\)[/tex]:
- [tex]\( y = 2.0 \)[/tex]
- So, [tex]\((x, y) = (5.0, 2.0)\)[/tex]

Here is the completed table:

[tex]\[ \begin{tabular}{|c|c|c|} \hline $x$ & $y$ & $(x, y)$ \\ \hline $10+(-5)$ & $-1+3$ & (5, 2) \\ \hline $0-(-4)$ & $-6-(-6)$ & (4, 0) \\ \hline $-2 \times (-2)$ & $-1 \times 2$ & (4, -2) \\ \hline $-30 \div (-5)$ & $-18 \div 6$ & (6.0, -3.0) \\ \hline $-2+9$ & $3+(-5)$ & (7, -2) \\ \hline $-1-(-9)$ & $5-5$ & (8, 0) \\ \hline $-7 \times (-1)$ & $-2 \times (-1)$ & (7, 2) \\ \hline $-40 \div (-8)$ & $-10 \div (-5)$ & (5.0, 2.0) \\ \hline \end{tabular} \][/tex]

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