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The picture to the right illustrates 3 / 4 of a cake. Fractions can be used to represent ratios and division. [1] Thus the fraction 3 / 4 can be used to represent the ratio 3:4 (the ratio of the part to the whole), and the division 3 ÷ 4 (three divided by four).
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
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. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.
If a mixture contains substances A, B, C and D in the ratio 5:9:4:2 then there are 5 parts of A for every 9 parts of B, 4 parts of C and 2 parts of D. As 5+9+4+2=20, the total mixture contains 5/20 of A (5 parts out of 20), 9/20 of B, 4/20 of C, and 2/20 of D. If we divide all numbers by the total and multiply by 100, we have converted to ...
The first four partial sums of the series 1 + 2 + 3 + 4 + ⋯.The parabola is their smoothed asymptote; its y-intercept is −1/12. [1]The infinite series whose terms ...
For example, given that there is a pattern of odds of 5/4, 7/4, 9/4 and so on, odds which are mathematically 3/2 are more easily compared if expressed in the equivalent form 6/4. Fractional odds are also known as British odds, UK odds, [9] or, in that country, traditional odds. They are typically represented with a "/" but can also be ...
Slices of approximately 1/8 of a pizza. A unit fraction is a positive fraction with one as its numerator, 1/ n.It is the multiplicative inverse (reciprocal) of the denominator of the fraction, which must be a positive natural number.
For example, when d=4, the hash table for two occurrences of d would contain the key-value pair 8 and 4+4, and the one for three occurrences, the key-value pair 2 and (4+4)/4 (strings shown in bold). The task is then reduced to recursively computing these hash tables for increasing n , starting from n=1 and continuing up to e.g. n=4.