Certainly! Let's solve the problem step by step to find the missing digits in the given subtraction equation:
[tex]\[
\begin{array}{r}
\square 2,048 \\
-12,218 \\
\hline 69,8 \square 0
\end{array}
\][/tex]
We will start by analyzing the columns from right to left:
1. Column of units (rightmost):
[tex]\(8 - 8 = 0\)[/tex]
No missing digit here.
2. Column of tens:
[tex]\(\square - 1 = \square\)[/tex]
However, the result is [tex]\(8\)[/tex]. Notice we won't directly solve this yet but use it in reverse conjunction.
3. Column of hundreds:
[tex]\(0 - 2 = 8\)[/tex]
This implies borrowing took place because [tex]\(0\)[/tex] is less than [tex]\(2\)[/tex]. Actually, [tex]\(10 - 2 = 8\)[/tex] with a borrow from the next left.
4. Column of thousands:
[tex]\(\square - 2 = 9\)[/tex]
This operation is generally acknowledging [tex]\( \square \)[/tex] was originally one extra due to the borrow we took. Simplified:
Let's denote the original being [tex]\( \square - 1 = 6 \Rightarrow \square = 7\)[/tex].
Lastly, our leading column:
[tex]\(\square - 2 = 6 \Rightarrow \)[/tex] Borrow again we consider [tex]\( \square = 7 + 1 = 9\)[/tex].
So the missing digits filled here should form a complete fit:
Thus, [tex]\(\square\)[/tex] value is [tex]\(\boxed{9}\)[/tex].
Revised: [tex]\(\square = 2\)[/tex].
Combining leads us to conclude missing leading must have identified.
Final Answer would be [tex]\( (9,2)\)[/tex].
