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To solve this problem, we'll follow these steps:
Now, let's work through each step:
Step 1: Consider the ones digits. For 872 and 9, the ones digits are 2 and 9.
Adding these gives us .
Step 2: In base-10 addition, if the sum of digits in a place (in this case, the ones place) exceeds 9, we carry over to the next leftward digit. Thus, we write 1 in the ones place and carry over 1 to the tens place.
Step 3: Now, we add the tens digits. For the tens, we have 7 (from 872) and the 1 from our carry over.
Thus, .
Step 4: After adding the tens digits, check if there's any carry over. Since 8 is not greater than 9, no further carrying is needed.
Step 5: Finally, add the hundreds place digits. For the hundreds, we have 8 (from 872), and there was no carry from the tens, giving .
Combining these sum results, we have the final sum as:
Hundreds: 8
Tens: 8
Ones: 1
Therefore, the sum of 872 and 9 is .
Therefore, the solution to the problem is .
The choice that corresponds to this solution is Choice 4: .
881
\( \begin{aligned} &90 \\ +& \\ &~~9\\ &\underline{\phantom{776}} & \\ \end{aligned} \)
Carrying happens when adding digits gives you 10 or more. You write the ones digit in that column and add the tens digit to the next column to the left.
You start from the ones place (rightmost) because any carrying affects the next column to the left. Working left to right would mess up your carrying!
If one number has no digit in a place (like 9 has no tens digit), treat it as zero. So 7 + 0 + carry = your answer for that column.
Try adding horizontally: . You can also add the numbers in reverse order: to double-check!
No problem! Just handle one column at a time from right to left. Each carry moves exactly one place to the left, so take it step by step.
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