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Q: How do you regroup 39 by using tens and ones?

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4 tens + 9 ones 3 tens + 19 ones 2 tens + 29 ones 1 ten + 39 ones 49 ones.

39 tens is 390.

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

There are 39 tens in 390. 390/10 = 39

39 tens.

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

40

3900/10=390

39

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.

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