To find how many 7s are in the product of 357 and 49, we first calculate the product: ( 357 \times 49 = 17,493 ). Next, we need to determine how many times 7 divides evenly into 17,493. Dividing, we find that ( 17,493 \div 7 = 2,498.71 ), which means there are 2 complete 7s in the product. Thus, the product contains 2 full 7s.
The answer is 7s + 49.
5.28751 7s equals 37.
To find how many 7s are in 93, you can divide 93 by 7. When you do this calculation, 93 ÷ 7 equals approximately 13.29. This means there are 13 full 7s in 93, with a remainder.
7,14,21,28,35
To find how many 7s are in the product of 357 and 49, we first calculate the product: ( 357 \times 49 = 17,493 ). Next, we need to determine how many times 7 divides evenly into 17,493. Dividing, we find that ( 17,493 \div 7 = 2,498.71 ), which means there are 2 complete 7s in the product. Thus, the product contains 2 full 7s.
The answer is 7s + 49.
Expressed algebraically, this would equal 7s - 5.
5.28751 7s equals 37.
To determine how many times the digit 7 appears in the number 49, we need to break down the number into its individual digits. In this case, 49 has two digits: 4 and 9. Since there is no 7 in either of these digits, the number 49 does not contain the digit 7. Therefore, there are zero instances of the digit 7 in the number 49.
To find how many 7s are in 93, you can divide 93 by 7. When you do this calculation, 93 ÷ 7 equals approximately 13.29. This means there are 13 full 7s in 93, with a remainder.
7,14,21,28,35
5
14
18
80.
6