'√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT]

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'√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT]

02-18-2020, 07:38 PM (This post was last modified: 02-18-2020 07:48 PM by Massimo Gnerucci.)

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'√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-18-2020, 06:49 PM

≈e [NT] - Massimo Gnerucci - 02-18-2020 07:38 PM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Paul Dale - 02-18-2020, 10:00 PM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-18-2020, 10:13 PM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Paul Dale - 02-19-2020, 02:51 AM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-19-2020, 04:36 AM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Dave Britten - 02-19-2020, 12:52 PM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - BruceH - 02-19-2020, 09:24 PM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - mfleming - 02-20-2020, 01:36 AM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-19-2020, 06:34 PM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Paul Dale - 02-20-2020, 12:36 AM

RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-20-2020, 07:29 PM

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