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This is an addition problem in base-5 (quinary system), where numbers are represented using digits 0 through 4. The task is to add 44 (base 5) and 43 (base 5).

Let me break it down step by step:

### Step 1: Convert the numbers from base 5 to base 10.

- \( 44_5 \): 
   - \( 4 \times 5^1 + 4 \times 5^0 = 4 \times 5 + 4 \times 1 = 20 + 4 = 24 \)

- \( 43_5 \): 
   - \( 4 \times 5^1 + 3 \times 5^0 = 4 \times 5 + 3 \times 1 = 20 + 3 = 23 \)

### Step 2: Add the base-10 equivalents.

\( 24 + 23 = 47 \)

### Step 3: Convert the sum (47) back to base 5.

To convert 47 (base 10) to base 5:
- Divide 47 by 5:
  - \( 47 \div 5 = 9 \) remainder \( 2 \)
  - \( 9 \div 5 = 1 \) remainder \( 4 \)
  - \( 1 \div 5 = 0 \) remainder \( 1 \)

So, \( 47_{10} = 142_5 \).

### Final Answer:
The sum of \( 44_5 + 43_5 \) is \( 142_5 \).

Would you like further clarification or more details? Here are five related questions to explore:

1. How do you subtract numbers in base-5?
2. How would you multiply numbers in base-5?
3. Can you explain the concept of other bases like base-8 (octal) or base-2 (binary)?
4. What are some real-world applications of different base systems?
5. How would you verify the result of this base-5 addition?

**Tip:** When adding numbers in other bases, it's often easier to first convert them to base-10, perform the operation, and then convert back to the desired base.

What is the sum of 44 (base 5) and 43 (base 5)?

This page provides a detailed step-by-step solution to adding the numbers 44 and 43 in base-5, explaining how to convert between base-5 and base-10 and perform the addition. The final answer is 142 in base-5. Suitable for students in Grades 5-7 learning about number systems.

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