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.
7,14,21,28,35
6
8
5
10
6
answer:4
14