one 10
13789
hundred thousands,ten thousands,thousands,hundreds,tens,ones
How to get 739,652 as the answer what is the number that has 652 in the ones period and 739 in the thousands period? to the right of the comma is ones period to the left is thousands period
9 hundred thousands 15 ten thousands 6 thousands 3 hundreds 7 ones in standard form is: 921,307
EXAMPLE: 24+2= 26 NO REGROUPING 56+3=59 NO REGROUPING 24+8=32 IS REGROUPING 56+4=60 IS REGROUPING TAKING THE ONES PLACE ONLY: FIRST EXAMPLE 4+2=6 HAS TO BE LESS THAN 9 4+8=12 YOU MAKE 10 IN THE ONES PLACE YOU CARRY OVER WHICH NOW THEY ARE CALLING REGROUPING. WE JUST CALLED IT CARRYING OVER AND BORROWING. HOPE THIS HELPS.
How much does 147 thousands equal in millions
one 10
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. This problem might be easier to visualize if you copy it vertically. 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
9, 29, 39, 49,59,69,79,89,99.............etc
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: 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
13789
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 is no four digit number where the ones is twice the tens, the hundreds is five less than the ones, and the thousands is the sum of the tens and hundreds. int ones, tens, hundreds, thousands; for (thousands=1; thousands<10; thousands++) { /**/ for (hundreds=0; hundreds<10; hundreds++) { /**/ /**/ for (tens=0; tens<10; tens++) { /**/ /**/ /**/ for (ones=0; ones<10; ones++) { /**/ /**/ /**/ /**/ if (ones != 2 * tens) break; /**/ /**/ /**/ /**/ if (hundreds != ones - 5) break; /**/ /**/ /**/ /**/ if (thousands != tens + hundreds) break; /**/ /**/ /**/ /**/ printf ("dd\n", thousands, hundreds, tens, ones); /**/ /**/ /**/ } /**/ /**/ } /**/ } }
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
212300 ones
900 is 0.9 thousands.