Question Number 190746 by leicianocosta last updated on 10/Apr/23 Answered by Mathspace last updated on 12/Apr/23 $$\left.{a}\right)\int_{\mathrm{1}} ^{\mathrm{4}} {x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{4}} \left(\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\…
Question Number 125202 by bemath last updated on 09/Dec/20 $$\:{Solve}\:{the}\:{reccurence}\:{relation} \\ $$$${a}_{{n}} \:=\:\mathrm{2}\left({a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \right)\:;\:{given}\:{a}_{\mathrm{0}} =\mathrm{1}\: \\ $$$${and}\:{a}_{\mathrm{1}} =\:\mathrm{0}. \\ $$ Answered by liberty last…
Question Number 125186 by Mathgreat last updated on 08/Dec/20 Answered by bemath last updated on 08/Dec/20 $$\left(\mathrm{1}\right)\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{9}}\right)\:= \\ $$$$\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{27}}}\right)=\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{12}}{\mathrm{26}}\right)=\:\mathrm{arctan}\:\left(\frac{\mathrm{6}}{\mathrm{13}}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{6}}{\mathrm{13}}\right)+\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{7}}{\mathrm{19}}\right)= \\ $$$$\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\frac{\mathrm{6}}{\mathrm{13}}+\frac{\mathrm{7}}{\mathrm{19}}}{\mathrm{1}−\frac{\mathrm{6}}{\mathrm{13}}.\frac{\mathrm{7}}{\mathrm{19}}}\right)\:=\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{205}}{\mathrm{205}}\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$…
Question Number 125185 by bemath last updated on 08/Dec/20 $$\:\begin{cases}{\sqrt{{x}}\:+\:{y}\:=\:\mathrm{11}}\\{{x}\:+\:\sqrt{{y}}\:=\:\mathrm{7}\:}\end{cases} \\ $$ Answered by Olaf last updated on 09/Dec/20 $$ \\ $$$$\begin{cases}{\sqrt{{x}}+{y}\:=\:\mathrm{11}\:\left(\mathrm{1}\right)}\\{{x}+\sqrt{{y}}\:=\:\mathrm{7}\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{4},\mathrm{9}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{quite}\:\mathrm{evident}\:\mathrm{solution}. \\…
Question Number 190705 by Ari last updated on 09/Apr/23 Answered by a.lgnaoui last updated on 09/Apr/23 $$\:\: \\ $$$$\mathrm{X}=+\mathrm{5} \\ $$$$ \\ $$$$\mathrm{Y}=+\mathrm{5} \\ $$$$…
Question Number 125169 by TANMAY PANACEA last updated on 08/Dec/20 $${algebra} \\ $$ Commented by TANMAY PANACEA last updated on 08/Dec/20 $$ \\ $$ Commented…
Question Number 190704 by mathlove last updated on 09/Apr/23 $${f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+{x}\centerdot{y} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{10}\:\:{finde}\:{f}\left(\mathrm{2022}\right)=? \\ $$ Answered by mahdipoor last updated on 09/Apr/23 $${f}\left({n}+\mathrm{1}\right)=\left[{f}\left({n}\right)\right]+{f}\left(\mathrm{1}\right)+{n}= \\ $$$$\left[{f}\left({n}−\mathrm{1}\right)+{f}\left(\mathrm{1}\right)+\left({n}−\mathrm{1}\right)\right]+{f}\left(\mathrm{1}\right)+{n}=… \\…
Question Number 59615 by hhghg last updated on 12/May/19 $$\mathrm{6}+\left(\frac{\mathrm{1}}{\mathrm{5}}×\mathrm{7}\right) \\ $$ Answered by Forkum Michael Choungong last updated on 12/May/19 $$=\frac{\mathrm{6}}{\mathrm{1}}+\frac{\mathrm{7}}{\mathrm{5}} \\ $$$$=\frac{\mathrm{30}+\mathrm{7}}{\mathrm{5}} \\…
Question Number 59614 by hhghg last updated on 12/May/19 $$\mathrm{1}\frac{\mathrm{1}}{\mathrm{7}}+\mathrm{1}\frac{\mathrm{1}}{\mathrm{14}} \\ $$ Answered by Forkum Michael Choungong last updated on 12/May/19 $$=\:\mathrm{2}\frac{\mathrm{2}+\mathrm{1}}{\mathrm{14}} \\ $$$$=\:\mathrm{2}\frac{\mathrm{3}}{\mathrm{14}} \\…
Question Number 190677 by Shrinava last updated on 08/Apr/23 $$\mathrm{If}\:\:\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\:\:\:\:\:\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,\mathrm{a}_{\mathrm{3}} \:,\:\mathrm{b}_{\mathrm{1}} ,\mathrm{b}_{\mathrm{2}} ,\mathrm{b}_{\mathrm{3}} \:>\:\mathrm{0} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{a}_{\mathrm{1}} ^{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} \:\mathrm{b}_{\mathrm{1}}…