65
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
Numbers that would require regrouping in the ones place when added to 21 are any numbers from 9 to 19. When adding a number in this range to 21, the sum will exceed 30, necessitating regrouping in the ones place to carry over to the tens place. For example, when adding 19 to 21, the sum is 40, requiring regrouping.
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
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
In 3 ten thousands, there are 30,000. To find the number of ones in this figure, you can break it down: 30,000 contains 30 thousands, which is equivalent to 30,000 ones. Therefore, there are 30,000 ones in 3 ten thousands.
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); /**/ /**/ /**/ } /**/ /**/ } /**/ } }
To determine how many thousands equal 9000 ones, you need to understand the concept of place value. In the number 9000, the 9 is in the thousands place, meaning there are 9 thousands. Each thousand is equal to 1000 ones, so 9000 ones is equal to 9 thousands.
To determine how many ones are in 8 ten thousands, you can first convert ten thousands to ones. Since one ten thousand equals 10,000, multiplying 8 ten thousands by 10,000 gives you 80,000. Therefore, there are 80,000 ones in 8 ten thousands.