Question Number 149424 by mathdanisur last updated on 05/Aug/21 $${geometric}\:{series}\:\:\frac{{b}_{\mathrm{4}} \centerdot{b}_{\mathrm{7}} \centerdot{b}_{\mathrm{10}} }{{b}_{\mathrm{1}} \centerdot{b}_{\mathrm{3}} \centerdot{b}_{\mathrm{5}} }\:=\:\mathrm{2}^{\mathrm{12}} \\ $$$${find}\:\:\:\frac{{b}_{\mathrm{5}} }{{b}_{\mathrm{2}} }\:=\:? \\ $$ Answered by nimnim…
Question Number 149423 by mathdanisur last updated on 05/Aug/21 $$\sqrt[{\mathrm{8}}]{\mathrm{2207}\:-\:\frac{\mathrm{1}}{\mathrm{2207}\:-\:\frac{\mathrm{1}}{\mathrm{2207}\:-\:…}}}\:\:=\:? \\ $$ Answered by dumitrel last updated on 05/Aug/21 $${x}=\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−….}}\Rightarrow{x}=\mathrm{2207}−\frac{\mathrm{1}}{{x}}\Rightarrow{x}=\frac{\mathrm{2207}+\mathrm{21}\centerdot\mathrm{47}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{16}} =…=\frac{\mathrm{2}^{\mathrm{15}} \left(\mathrm{2207}+\mathrm{21}\centerdot\mathrm{47}\sqrt{\mathrm{5}}\right)}{\mathrm{2}^{\mathrm{16}} }={x}\Rightarrow\sqrt[{\mathrm{8}}]{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{16}}…
Question Number 83874 by Power last updated on 07/Mar/20 Commented by niroj last updated on 07/Mar/20 $$\:\:\:\mathrm{xy}\left(\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)=\:\mathrm{24}….\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{10}……\left(\mathrm{ii}\right) \\ $$$$\:\:\mathrm{multipy}\:\mathrm{by}\:\mathrm{xy}\:\mathrm{in}\:\left(\mathrm{ii}\right)\mathrm{then}\:\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\:…
Question Number 149410 by mathdanisur last updated on 05/Aug/21 $$\frac{{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{20}} \:+\:{log}_{\mathrm{2}} \:\mathrm{20}\:\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:-\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{5}} }{{log}_{\mathrm{2}} \:\mathrm{20}\:+\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 149408 by mathdanisur last updated on 05/Aug/21 $${cos}\left(\frac{\pi}{\mathrm{2}}\:-\:\frac{\mathrm{1}}{\mathrm{2}}\:{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}\right)\:=\:? \\ $$ Commented by liberty last updated on 05/Aug/21 $$\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{sin}\:\mathrm{u} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{u}\:\Rightarrow\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{2u} \\ $$$$\mathrm{cos}\:\mathrm{2u}=\frac{\mathrm{4}}{\mathrm{5}}\Rightarrow\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{u}=\frac{\mathrm{4}}{\mathrm{5}}…
Question Number 83871 by jagoll last updated on 07/Mar/20 $$\mathrm{If}\:\mathrm{equation}\: \\ $$$$\begin{cases}{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{10}}\\{\mathrm{x}+\mathrm{2y}=\:\mathrm{5z}}\end{cases} \\ $$$$\mathrm{has}\:\mathrm{solution}\:\mathrm{is}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right).\: \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{2b}+\mathrm{3c}\: \\…
Question Number 149405 by mathdanisur last updated on 05/Aug/21 $${if}\:\:\:\int{f}\left({x}\right){dx}\:=\:{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −{c} \\ $$$${find}\:\:\:{f}\left(−\mathrm{2}\right)\:=\:? \\ $$ Commented by liberty last updated on 05/Aug/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3x}^{\mathrm{2}} −\mathrm{6x}…
Question Number 149400 by mathdanisur last updated on 05/Aug/21 Answered by liberty last updated on 05/Aug/21 $$\mathrm{let}\:\mathrm{tan}\:\mathrm{2}=\mathrm{x}\:\begin{cases}{\mathrm{sin}\:\mathrm{2}=\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}}\\{\mathrm{cos}\:\mathrm{2}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}}\end{cases} \\ $$$$\mathrm{let}\:\mathrm{cos}\:\mathrm{6}=\mathrm{y}\Rightarrow\mathrm{2cos}\:^{\mathrm{2}} \mathrm{3}−\mathrm{1}=\mathrm{y} \\ $$$$\mathrm{cos}\:\mathrm{3}=\sqrt{\frac{\mathrm{y}+\mathrm{1}}{\mathrm{2}}}\:\wedge\:\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{3}=\mathrm{y}…
Question Number 149395 by mathdanisur last updated on 05/Aug/21 Answered by Olaf_Thorendsen last updated on 05/Aug/21 $$\int_{{a}} ^{{b}} {f}\left({x}\right)\:{dx}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{b}−{a}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left({a}+{k}\frac{{b}−{a}}{{n}}\right) \\ $$$$\int_{\mathrm{3}} ^{\mathrm{5}}…
Question Number 83834 by Power last updated on 06/Mar/20 Commented by niroj last updated on 06/Mar/20 $$\:\:\int\:\frac{\:\:\mathrm{1}}{\:\sqrt{\mathrm{x}\left(\mathrm{a}−\mathrm{x}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{ax}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\:\int\:\frac{\:\mathrm{1}}{\:\sqrt{−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}.\frac{\mathrm{a}}{\mathrm{2}}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{−\left[\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}}…