(3*10)+(9*1)=39
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
39
Numbers that have 3 tens range from 30 to 39. This includes all the integers from 30 to 39, which totals 10 numbers: 30, 31, 32, 33, 34, 35, 36, 37, 38, and 39. Therefore, there are 10 numbers that have 3 tens.
To find out how many tens are in 384, divide 384 by 10, which gives you 38.4. This means there are 38 complete tens in 384. The next greatest number of tens would be 39.
4 tens + 9 ones 3 tens + 19 ones 2 tens + 29 ones 1 ten + 39 ones 49 ones.
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them horizontally. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
39 tens is 390.
There are 39 tens in 390. 390/10 = 39
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
39 tens.
39
The product of 39 multiplied by 39 is 1521. This can be calculated by multiplying the tens place of each number (3 x 3 = 9) to get the hundreds place of the result, and then multiplying each number by the other number's digits and adding them together (3 x 9 = 27) to get the tens and ones place of the result. Therefore, 39 x 39 = 1521.
Numbers that have 3 tens range from 30 to 39. This includes all the integers from 30 to 39, which totals 10 numbers: 30, 31, 32, 33, 34, 35, 36, 37, 38, and 39. Therefore, there are 10 numbers that have 3 tens.
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17
You need to regroup when you have something like 46-39. You would cross out the 6 and make it a 16, because you are taking ten from the 4, then you cross out the 4 then make it a three, then you should get 7.