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[0; 4, 4, 8, 16, 18, 5, 1, 1, 1, 1, 7, 1, 1, 6, 2, 9, 58, 1, 3, 4, …] [OEIS 100] Computed up to 1 011 597 392 terms by E. Weisstein. He also noted that while the Champernowne constant continued fraction contains sporadic large terms, the continued fraction of the Copeland–Erdős Constant do not exhibit this property. [Mw 85] Base 10 ...
In fractions like "2 nanometers per meter" (2 n m / m = 2 nano = 2×10 −9 = 2 ppb = 2 × 0.000 000 001), so the quotients are pure-number coefficients with positive values less than or equal to 1. When parts-per notations, including the percent symbol (%), are used in regular prose (as opposed to mathematical expressions), they are still pure ...
German mathematician Simon Jacob (d. 1564) noted that consecutive Fibonacci numbers converge to the golden ratio; [25] this was rediscovered by Johannes Kepler in 1608. [26] The first known decimal approximation of the (inverse) golden ratio was stated as "about 0.6180340 {\displaystyle 0.6180340} " in 1597 by Michael Maestlin of the ...
However the titles of bonds issued by governments and other issuers use the fractional form, e.g. "3 + 1 ⁄ 2 % Unsecured Loan Stock 2032 Series 2". (When interest rates are very low, the number 0 is included if the interest rate is less than 1%, e.g. "0 + 3 ⁄ 4 % Treasury Stock", not "3 ⁄ 4 % Treasury Stock".) It is also widely accepted ...
[1] [2] [3] Equal quotients correspond to equal ratios. A statement expressing the equality of two ratios is called a proportion . Consequently, a ratio may be considered as an ordered pair of numbers, a fraction with the first number in the numerator and the second in the denominator, or as the value denoted by this fraction.
The first 3 powers of 2 with all but last digit odd is 2 4 = 16, 2 5 = 32 and 2 9 = 512. The next such power of 2 of form 2 n should have n of at least 6 digits. The only powers of 2 with all digits distinct are 2 0 = 1 to 2 15 = 32 768, 2 20 = 1 048 576 and 2 29 = 536 870 912.
The reason is that 3 is a divisor of 9, 11 is a divisor of 99, 41 is a divisor of 99999, etc. To find the period of 1 / p , we can check whether the prime p divides some number 999...999 in which the number of digits divides p − 1. Since the period is never greater than p − 1, we can obtain this by calculating 10 p−1 − 1 / p ...
Propeller-driven airliners had useful load fractions on the order of 25–35%. Modern jet airliners have considerably higher useful load fractions, on the order of 45–55%. For orbital rockets the payload fraction is between 1% and 5%, while the useful load fraction is perhaps 90%.