To find the product of \(48\) and \(7,326\) using the place-value chart, we break down \(48\) into its place values:
- \(40\) (in the tens place)
- \(8\) (in the ones place)
Now, we will multiply each part of \(48\) by each part of \(7,326\), breaking \(7,326\) into its place values as well:
- \(7,000\)
- \(300\)
- \(20\)
- \(6\)
Now we'll set up the multiplication in a place-value chart:
| | Thousands (7,000) | Hundreds (300) | Tens (20) | Ones (6) | |---------------------|-------------------|----------------|-----------|----------| | Tens (40) | \(40 \times 7,000 = 280,000\) | \(40 \times 300 = 12,000\) | \(40 \times 20 = 800\) | \(40 \times 6 = 240\) | | Ones (8) | \(8 \times 7,000 = 56,000\) | \(8 \times 300 = 2,400\) | \(8 \times 20 = 160\) | \(8 \times 6 = 48\) |
Now we fill in the products:
| | Thousands (7,000) | Hundreds (300) | Tens (20) | Ones (6) | |---------------------|-------------------|----------------|-----------|----------| | Tens (40) | 280,000 | 12,000 | 800 | 240 | | Ones (8) | 56,000 | 2,400 | 160 | 48 |
Sum of Products
Now we can sum all the products:
-
From Tens (40):
- \(280,000\)
- \(12,000\)
- \(800\)
- \(240\)
- Total for Tens (40):
- \(280,000 + 12,000 + 800 + 240 = 292,040\)
-
From Ones (8):
- \(56,000\)
- \(2,400\)
- \(160\)
- \(48\)
- Total for Ones (8):
- \(56,000 + 2,400 + 160 + 48 = 58,608\)
Final Sum
Adding these two totals gives:
\[ 292,040 + 58,608 = 350,648 \]
Thus, \(48 \times 7,326 = 350,648\).
Relation to Standard Algorithm
The place-value multiplication method is directly related to the standard algorithm for multiplication, which involves breaking numbers into place values, multiplying each part separately, and then summing the results. The end result demonstrates that both methods lead to the same product, confirming that they are equivalent approaches to multiplication.
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