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Video converter Developer License Supported platform Windows Mac OS X Linux Any Video Converter: Anvsoft Inc. Freeware: Yes: Yes: No Avidemux: Mean, Gruntster, Fahr: GPL-2.0-or-later: Yes: Yes: Yes Dr. DivX: DivX, Inc. Adware bundled 15-day trial: Yes: Yes: No DVDVideoSoft Free Studio: DVDVideoSoft: Shareware (requires paid membership for basic ...
A list of articles about numbers (not about numerals). Topics include powers of ten, notable integers, prime and cardinal numbers, and the myriad system.
Fourth powers are also formed by multiplying a number by its cube. Furthermore, they are squares of squares. Some people refer to n 4 as n tesseracted, hypercubed, zenzizenzic, biquadrate or supercubed instead of “to the power of 4”. The sequence of fourth powers of integers, known as biquadrates or tesseractic numbers, is:
Freemake Video Converter 2.0 was a major update that integrated two new functions: ripping video from online portals and Blu-ray disc creation and burning. [13] [14] Version 2.1 implemented suggestions from users, including support for subtitles, ISO image creation, and DVD to DVD/Blu-ray conversion. [15]
The cube of a number n is denoted n 3, using a superscript 3, [a] for example 2 3 = 8. The cube operation can also be defined for any other mathematical expression, for example (x + 1) 3. The cube is also the number multiplied by its square: n 3 = n × n 2 = n × n × n. The cube function is the function x ↦ x 3 (often denoted y = x 3) that
To do this, he called the numbers up to a myriad myriad (10 8) "first numbers" and called 10 8 itself the "unit of the second numbers". Multiples of this unit then became the second numbers, up to this unit taken a myriad myriad times, 10 8 ·10 8 =10 16. This became the "unit of the third numbers", whose multiples were the third numbers, and ...
Because of the factorization (2n + 1)(n 2 + n + 1), it is impossible for a centered cube number to be a prime number. [3] The only centered cube numbers which are also the square numbers are 1 and 9, [4] [5] which can be shown by solving x 2 = y 3 + 3y, the only integer solutions being (x,y) from {(0,0), (1,2), (3,6), (12,42)}, By substituting a=(x-1)/2 and b=y/2, we obtain x^2=2y^3+3y^2+3y+1.