89 with remainder 1.
It goes: 357/8 = 44 times with a remainder of 5
119
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 determine how many times 7 can fit into 17,493 by dividing: (17,493 \div 7 = 2,499). Therefore, there are 2,499 instances of 7 in the product 357 and 49.
To find the partial products for the multiplication of 357 and 48, you can break down the numbers. For instance, you can express 48 as 40 + 8. Then, multiply 357 by each part: (357 \times 40 = 14,280) and (357 \times 8 = 2,856). The partial products are 14,280 and 2,856.
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.
3 times
It goes: 357/8 = 44 times with a remainder of 5
It is: 357/8 = 44 with a remainder of 5
89.25 times.
21 times.
To determine how many times 4 can go into 357, you would perform long division. The quotient is the result of dividing the dividend (357) by the divisor (4). In this case, 4 can go into 357 approximately 89 times, with a remainder of 1.
119
340 = 20 x 17 357 - 340 = 17 17 times with 17 remaining
357 cannot go into 5 at all, as 357 is greater than 5. In mathematical terms, 5 divided by 357 results in a fraction less than 1. Therefore, 357 goes into 5 zero times.
18 times.
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 determine how many times 7 can fit into 17,493 by dividing: (17,493 \div 7 = 2,499). Therefore, there are 2,499 instances of 7 in the product 357 and 49.
44 times with 5 left over