Question Number 209357 by mnjuly1970 last updated on 08/Jul/24 $$ \\ $$$$\:\:\:\:\:\:{Evaluate}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{B}_{{n}} =\:\underset{{k}=\mathrm{3}} {\overset{{n}} {\prod}}\:\frac{\:{k}^{\:\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} \:+\:{k}\:−\mathrm{6}}=\:? \\ $$ Answered by…
Question Number 209385 by efronzo1 last updated on 08/Jul/24 Answered by Rasheed.Sindhi last updated on 08/Jul/24 $${p}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+\mathrm{1}\Rightarrow{p}\left(\mathrm{1}\right)={a}+{b}+\mathrm{1} \\ $$$${q}\left({x}\right)={bx}^{\mathrm{2}} +{ax}+\mathrm{1}\Rightarrow{q}\left(\mathrm{1}\right)={b}+{a}+\mathrm{1} \\ $$$${p}\left(\:{q}\left(\mathrm{1}\right)\:\right)={q}\left(\:{p}\left(\mathrm{1}\right)\:\right) \\ $$$${p}\left({a}+{b}+\mathrm{1}\right)={q}\left({a}+{b}+\mathrm{1}\right)…
Question Number 209309 by hardmath last updated on 06/Jul/24 $$\mathrm{m}\:,\:\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{m}\:\geqslant\:\mathrm{2}\:\:\:\mathrm{and}\:\:\:\mathrm{n}\:\geqslant\:\mathrm{2} \\ $$$$\mathrm{p}\:>\:\mathrm{0}\:\:\:\mathrm{and}\:\:\:\mathrm{q}\:>\:\mathrm{0} \\ $$$$\mathrm{p}\:+\:\mathrm{q}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\left(\mathrm{1}−\mathrm{q}^{\boldsymbol{\mathrm{n}}} \right)^{\boldsymbol{\mathrm{m}}} \:+\:\left(\mathrm{1}−\mathrm{p}^{\boldsymbol{\mathrm{m}}} \right)^{\boldsymbol{\mathrm{n}}} \:\geqslant\:\mathrm{1} \\ $$ Terms…
Question Number 209290 by klipto last updated on 06/Jul/24 $$\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}−^{\mathrm{1000}} \sqrt{\left(\mathrm{1}+\mathrm{8000}\boldsymbol{\mathrm{a}}\right)}=\mathrm{1000} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{a}} \\ $$ Commented by mr W last updated on 06/Jul/24 $${you}\:{can}\:{only}\:{approximate}!…
Question Number 209318 by essaad last updated on 06/Jul/24 Answered by Frix last updated on 06/Jul/24 $$\mathrm{We}\:\mathrm{know}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{k}}\:=\infty \\ $$$${k}>\mathrm{1}:\:\frac{\mathrm{1}}{\:\sqrt{{k}}}>\frac{\mathrm{1}}{\:{k}}\:\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\:\sqrt{{k}}}\:>\underset{{k}=\mathrm{1}} {\overset{\infty}…
Question Number 209240 by Jubr last updated on 05/Jul/24 $${If}\:\:{x}\:\:+\:\:\frac{\mathrm{49}}{{x}\:+\:\mathrm{48}}\:\:=\:\:−\:\mathrm{34} \\ $$$${find}\:\:\left(\mathrm{2}{x}\:+\:\mathrm{83}\right)^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\left(\mathrm{2}{x}\:+\:\mathrm{83}\right)^{\mathrm{3}} } \\ $$ Answered by som(math1967) last updated on 05/Jul/24 $${x}+\mathrm{48}+\frac{\mathrm{49}}{{x}+\mathrm{48}}=−\mathrm{34}+\mathrm{48} \\…
Question Number 209263 by hardmath last updated on 05/Jul/24 $$ \\ $$6 different letters were written to 6 different people and 6 different envelopes were…
Question Number 209223 by Tawa11 last updated on 04/Jul/24 Answered by efronzo1 last updated on 04/Jul/24 $$\:\:\:\:\cancel{\underbrace{\Subset}} \\ $$ Commented by Tawa11 last updated on…
Question Number 209234 by Tawa11 last updated on 04/Jul/24 $$\mathrm{Arrange}\:\mathrm{in}\:\mathrm{descending}\:\mathrm{order}: \\ $$$$\:\:\:\:\sqrt{\mathrm{5}}\:\:−\:\:\sqrt{\mathrm{2}},\:\:\:\:\:\sqrt{\mathrm{7}}\:\:−\:\:\sqrt{\mathrm{5}}\:,\:\:\:\sqrt{\mathrm{13}}\:\:−\:\:\sqrt{\mathrm{11}}\:,\:\:\:\sqrt{\mathrm{19}}\:\:−\:\:\sqrt{\mathrm{17}} \\ $$ Answered by A5T last updated on 04/Jul/24 $${Same}\:{arrangement}: \\ $$$$\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}\left(>\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}\right)>\sqrt{\mathrm{7}}−\sqrt{\mathrm{5}}>\sqrt{\mathrm{13}}−\sqrt{\mathrm{11}}>\sqrt{\mathrm{19}}−\sqrt{\mathrm{17}} \\…
Question Number 209187 by mnjuly1970 last updated on 03/Jul/24 $$ \\ $$$$\:\:\:::\:\:\:\alpha\:,\:\beta\:\:{and}\:\:\gamma\:\:{are}\:{roots}\:{of}\:{the} \\ $$$$\:\:\:\:\:{following}\:\:{equation}\:.\:{Find}\:{the} \\ $$$$\:\:\:\:\:{value}\:\:{of}\:\:\:''\:\:\mathrm{F}\:\:''\::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{E}{quation}\::\:\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{2}{x}\:\:−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{F}\::=\:\alpha^{\:\mathrm{5}} \:+\:\beta^{\:\mathrm{5}} \:+\:\gamma^{\:\mathrm{5}}…