To fill in Liam’s missing work while solving \( 12 \times 14 \), we can break it down step-by-step:
- Set up the multiplication:
12
× 14
-
Multiply the ones place:
- 4 (from 14) multiplied by 2 (the ones place of 12): \[ 4 \times 2 = 8 \]
-
Multiply the tens place (4 again):
- 4 multiplied by 1 (the tens place of 12): \[ 4 \times 1 = 4 \] Because we are in the tens place, we write this as \( 40 \) (just put a zero).
So the first row of the addition for \( 12 \times 4 \) will be:
8 (from 2 × 4)
+ 40 (from 1 × 4)
-----
48 (this would actually be added separately)
- Move to the next row for the tens place:
- Now multiply 10 (from 12, which is 1 in the tens place multiplied by 14):
10 (actually 1 × 14)
+ 120 (since we'll effectively be doing this, adding that 1 from moving over)
- 1 multiplied by 4 again in the tens adds no new work, so we concatenate as it is.
Putting it all together in the addition step:
8
+ 40
+ 120
-----
168 (final result after all additions)
So, filling in Liam's table:
1 2
× 1 4
-----------------
8
+ 4 0
+ 1 2 0
-----------------
1 6 8 (Final answer)
Thus, the digits to fill in would be: 1st Box: 8 2nd Box: 4 3rd Box: 0 4th Box: 2 5th Box: 1 (Since we have 12 in tens) Bottom Box: 6 (Final result at the bottom)
So the answer to fill in the boxes appropriately is:
8
+ 40
+ 120
Final answer is that last row would result in 1 6 8 after all proper additions done correctly.
1 answer