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In the second step, they were divided by 3. The final result, 4 / 3 , is an irreducible fraction because 4 and 3 have no common factors other than 1. The original fraction could have also been reduced in a single step by using the greatest common divisor of 90 and 120, which is 30. As 120 ÷ 30 = 4, and 90 ÷ 30 = 3, one gets
The lowest common denominator of a set of fractions is the lowest number that is a multiple of all the ... 8 because each measure is divided by 4 and by 3, ...
For example, is not in lowest terms because both 3 and 9 can be exactly divided by 3. In contrast, is in lowest terms—the only positive integer that goes into both 3 and 8 evenly is 1. Using these rules, we can show that 5 / 10 = 1 / 2 = 10 / 20 = 50 / 100 , for example.
For example, 20 apples divide into five groups of four apples, meaning that "twenty divided by five is equal to four". This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient.
One is going to increase this place by using the number one borrowed from the column to the left. Therefore: 10 − 8 = 2. It is 10 rather than 0, because one borrowed from the Thousands place. 75 > 44 so no need to borrow, say "two hundred" Tens: 7 − 4 = 3, 5 > 4, so 5 - 4 = 1 Hence, the result is 2231.
A multiple of a number is the product of that number and an integer. For example, 10 is a multiple of 5 because 5 × 2 = 10, so 10 is divisible by 5 and 2. Because 10 is the smallest positive integer that is divisible by both 5 and 2, it is the least common multiple of 5 and 2.
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
Here, we set up a partial fraction for each descending power of the denominator. Then we solve for the numerators, A and B. As ( 1 − 2 x ) {\displaystyle (1-2x)} is a repeated factor, we now need to find two numbers, as so we need an additional relation in order to solve for both.