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Author: Tinku Tara

and-are-roots-of-the-following-equation-Find-the-value-of-F-Equation-x-3-2x-1-0-F-5-5-

Question Number 209187 by mnjuly1970 last updated on 03/Jul/24 $$ \\ $$$$\:\:\:::\:\:\:\alpha\:,\:\beta\:\:{and}\:\:\gamma\:\:{are}\:{roots}\:{of}\:{the} \\ $$$$\:\:\:\:\:{following}\:\:{equation}\:.\:{Find}\:{the} \\ $$$$\:\:\:\:\:{value}\:\:{of}\:\:\:''\:\:\mathrm{F}\:\:''\::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{E}{quation}\::\:\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{2}{x}\:\:−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{F}\::=\:\alpha^{\:\mathrm{5}} \:+\:\beta^{\:\mathrm{5}} \:+\:\gamma^{\:\mathrm{5}}…

Question-209116

Question Number 209116 by alcohol last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $${WLOG},\:{let}\:{a}\geqslant{b}\geqslant{c} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\leqslant\frac{\mathrm{3}}{{c}}\Rightarrow{c}\leqslant\mathrm{3} \\ $$$${when}\:{c}=\mathrm{3};\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\leqslant\frac{\mathrm{2}}{{b}}\Rightarrow{b}\leqslant\mathrm{3} \\ $$$${b}=\mathrm{3}\Rightarrow{a}=\mathrm{3};\:{b}=\mathrm{2}\Rightarrow{a}=\mathrm{6}\:\:\:\rightarrow\leftarrow \\ $$$$\Rightarrow\left({a},{b},{c}\right)=\left(\mathrm{3},\mathrm{3},\mathrm{3}\right)…

please-convert-2531-5000-to-base-5002-thanks-

Question Number 209167 by lmcp1203 last updated on 02/Jul/24 $${please}\:{convert}\:\:\mathrm{2531}_{\left(\mathrm{5000}\right)\:} {to}\:\:{base}\:\mathrm{5002}.\:\:{thanks}.\:\: \\ $$ Answered by mr W last updated on 03/Jul/24 $$\mathrm{2531}_{\left(\mathrm{5000}\right)} =\mathrm{2}×\mathrm{5000}^{\mathrm{3}} +\mathrm{5}×\mathrm{5000}^{\mathrm{2}} +\mathrm{3}×\mathrm{5000}+\mathrm{1}…

Question-209128

Question Number 209128 by Spillover last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $${The}\:{length}\:{of}\:{the}\:{diagonal}\:{of}\:{the}\:{square}=\mathrm{2}{r} \\ $$$$\Rightarrow{s}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \Rightarrow\mathrm{2}{s}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \Rightarrow{s}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}}…