Fill in the boxes.

1. 6 more than [tex]$-3$[/tex] = [tex]$\square$[/tex]

2. 18 less than [tex]$10$[/tex] = [tex]$\square$[/tex]

3. [tex]$(-20) \times (-28)$[/tex] = [tex]$\square$[/tex]

4. Additive inverse of 0 is [tex]$\square$[/tex]

5. [tex]$(-2) \times (-2)$[/tex] = [tex]$\square$[/tex]



Answer :

Certainly! Let's solve each problem step-by-step.

1. 6 more than [tex]\(-3\)[/tex]:
- Starting with [tex]\(-3\)[/tex]
- Adding 6 to it:
[tex]\(-3 + 6\)[/tex]
- Result: [tex]\(3\)[/tex]
- So, [tex]\(6\)[/tex] more than [tex]\(-3\)[/tex] is [tex]\(\boxed{3}\)[/tex]

3. 18 less than [tex]\(10\)[/tex]:
- Starting with [tex]\(10\)[/tex]
- Subtracting [tex]\(18\)[/tex] from it:
[tex]\(10 - 18\)[/tex]
- Result: [tex]\(-8\)[/tex]
- So, [tex]\(18\)[/tex] less than [tex]\(10\)[/tex] is [tex]\(\boxed{-8}\)[/tex]

5. [tex]\((-20) \times (-28)\)[/tex]:
- Evaluating the product of two negative numbers:
[tex]\((-20) \times (-28)\)[/tex]
- Result: [tex]\(560\)[/tex]
- So, [tex]\((-20) \times (-28)\)[/tex] is [tex]\(\boxed{560}\)[/tex]

7. Additive inverse of [tex]\(0\)[/tex]:
- The additive inverse of a number [tex]\(x\)[/tex] is a number that when added to [tex]\(x\)[/tex] yields 0.
- For [tex]\(0\)[/tex], the additive inverse is [tex]\(0\)[/tex] itself.
- So, the additive inverse of [tex]\(0\)[/tex] is [tex]\(\boxed{0}\)[/tex]

9. [tex]\((-2) \times (-2)\)[/tex]:
- Evaluating the product of two negative numbers:
[tex]\((-2) \times (-2)\)[/tex]
- Result: [tex]\(4\)[/tex]
- So, [tex]\((-2) \times (-2)\)[/tex] is [tex]\(\boxed{4}\)[/tex]

So, the filled boxes should be:
1. [tex]\(3\)[/tex]
3. [tex]\(-8\)[/tex]
5. [tex]\(560\)[/tex]
7. [tex]\(0\)[/tex]
9. [tex]\(4\)[/tex]

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