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8÷2(1+3)
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12-04-2023, 12:30 AM
Post: #12
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RE: 8÷2(1+3)
(12-03-2023 11:15 PM)dm319 Wrote: They are not equivalent to the multiplication operator IMO, though can be converted to it with brackets, i.e. 2a = (2 x a). If that is the case then: \( 2a^3 = (2 \times a)^3 = 2^3 \times a^3 = 8 \times a^3 \) Do you agree or do you come up with another special rule out of thin air? |
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RE: 8÷2(1+3) - rprosperi - 12-03-2023, 01:55 PM
RE: 8÷2(1+3) - Valentin Albillo - 12-03-2023, 02:43 PM
RE: 8÷2(1+3) - dm319 - 12-03-2023, 04:11 PM
RE: 8÷2(1+3) - dm319 - 12-03-2023, 06:50 PM
RE: 8÷2(1+3) - Johnh - 12-03-2023, 08:09 PM
No, never, not even once - striegel - 12-03-2023, 10:11 PM
RE: 8÷2(1+3) - Maximilian Hohmann - 12-03-2023, 10:25 PM
RE: 8÷2(1+3) - Thomas Klemm - 12-03-2023, 10:31 PM
RE: 8÷2(1+3) - dm319 - 12-03-2023, 11:15 PM
RE: 8÷2(1+3) - Thomas Klemm - 12-04-2023, 12:09 AM
RE: 8÷2(1+3) - Thomas Klemm - 12-04-2023 12:30 AM
RE: 8÷2(1+3) - Eddie W. Shore - 12-04-2023, 01:15 AM
RE: 8÷2(1+3) - John Garza (3665) - 12-04-2023, 06:42 AM
RE: 8÷2(1+3) - dm319 - 12-04-2023, 09:46 AM
RE: 8÷2(1+3) - toml_12953 - 12-06-2023, 09:42 AM
RE: 8÷2(1+3) - Steve Simpkin - 12-04-2023, 12:30 AM
RE: 8÷2(1+3) - klesl - 12-04-2023, 11:25 AM
RE: 8÷2(1+3) - johnb - 12-04-2023, 08:05 PM
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