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The entire fraction may be expressed as a single composition, in which case it is hyphenated, or as a number of fractions with a numerator of one, in which case they are not. (For example, two-fifths is the fraction 2 / 5 and two fifths is the same fraction understood as 2 instances of 1 / 5 .) Fractions should always be ...
43×5 = 215 Half of 3's neighbor is 0, plus 5 because 3 is odd, is 5. Half of 4's neighbor is 1. Half of the leading zero's neighbor is 2. 93×5=465 Half of 3's neighbor is 0, plus 5 because 3 is odd, is 5. Half of 9's neighbor is 1, plus 5 because 9 is odd, is 6. Half of the leading zero's neighbor is 4.
An extension to the rule of three was the double rule of three, which involved finding an unknown value where five rather than three other values are known. An example of such a problem might be If 6 builders can build 8 houses in 100 days, how many days would it take 10 builders to build 20 houses at the same rate? , and this can be set up as
When speaking the numbers in fractions, the spatial/chronological numbering system is used for denominators larger than 2 (2 as the denominator of a fraction is "half" rather than "second"), with a denominator of 4 sometimes spoken as "quarter" rather than "fourth". This system results in "two thirds" for 2 ⁄ 3 and "fifteen thirty-seconds ...
Twice (two times) as many as one. doubled. 100% Three 300% Two times more than one, twice more than one, or 200% more than one, because = +. Three times as many as one. tripled. 200% Four 400% Three times more than one, or 300% more than one, because = +. Four times as many as one. quadrupled. 300% Five 500%
In mathematics education, there was a debate on the issue of whether the operation of multiplication should be taught as being a form of repeated addition.Participants in the debate brought up multiple perspectives, including axioms of arithmetic, pedagogy, learning and instructional design, history of mathematics, philosophy of mathematics, and computer-based mathematics.
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If two divisions are done, a multiple of 5 · 5=25 rather than 5 must be used, because 25 can be divided by 5 twice. So the number of coconuts that could be in the pile is k · 25 – 4. k =1 yielding 21 is the smallest positive number that can be successively divided by 5 twice with remainder 1.