Question Number 149428 by mr W last updated on 06/Aug/21 Commented by mr W last updated on 06/Aug/21 $${three}\:{non}−{collinear}\:{points}\:{in}\:{space}\: \\ $$$${are}\:{given}\:{with}\:{coordinates} \\ $$$${A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}}…
Question Number 83892 by sahnaz last updated on 07/Mar/20 $$\int\frac{\mathrm{du}}{\mathrm{u}−\mathrm{u}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18357 by tawa tawa last updated on 19/Jul/17 $$\mathrm{From}\:\mathrm{the}\:\mathrm{topic}\:\mathrm{transformer} \\ $$$$ \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{e}\:=\:\sqrt{\mathrm{2}}\:\varepsilon\:\mathrm{cos}\left(\omega\mathrm{t}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 83893 by sahnaz last updated on 07/Mar/20 $$\int\frac{\mathrm{du}}{\mathrm{u}−\mathrm{u}^{\mathrm{2}} } \\ $$ Commented by abdomathmax last updated on 07/Mar/20 $$\int\:\:\frac{{du}}{{u}−{u}^{\mathrm{2}} }\:=\int\:\:\frac{{du}}{{u}\left(\mathrm{1}−{u}\right)}\:=\int\:\:\left(\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right){du} \\ $$$$={ln}\mid\frac{{u}}{\mathrm{1}−{u}}\mid\:+{C} \\…
Question Number 149425 by help last updated on 05/Aug/21 Answered by dumitrel last updated on 05/Aug/21 $${x}>\mathrm{0};{x}\neq\mathrm{1} \\ $$$$\sqrt[{\mathrm{4}}]{{log}_{\mathrm{3}} {x}}={y}>\mathrm{0}\Rightarrow{x}=\mathrm{3}^{{y}^{\mathrm{4}} } \\ $$$$\mathrm{9}^{{y}} −\mathrm{4}{x}^{\frac{\mathrm{1}}{{y}^{\mathrm{3}} }}…
Question Number 18355 by tawa tawa last updated on 19/Jul/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 149424 by mathdanisur last updated on 05/Aug/21 $${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 83891 by bagjamath last updated on 07/Mar/20 $$ \\ $$$$ \\ $$$$\mathrm{1111}^{\mathrm{2019}} \:\mathrm{mod}\:\mathrm{11111}=….? \\ $$ Commented by MJS last updated on 08/Mar/20 $$\mathrm{you}\:\mathrm{are}\:\mathrm{right},\:\mathrm{I}\:\mathrm{misread}\:\mathrm{1111}^{\mathrm{2019}}…
Question Number 83886 by mr W last updated on 07/Mar/20 $${To}\:{the}\:{developers}\:{of}\:{TinkuTara}: \\ $$$${problem}\:\mathrm{1}: \\ $$$${i}\:{get}\:{no}\:{notifications}\:{when}\:{my}\:{posts} \\ $$$${are}\:{updated}. \\ $$$$ \\ $$$${problem}\:\mathrm{2}: \\ $$$${i}\:{can}\:{edit}\:{my}\:{post},\:{see}\:{picture}\:\mathrm{1},\:{but} \\ $$$${the}\:{content}\:{is}\:{not}\:{visiable},\:{see}\:{picture}\:\mathrm{2}.…
Question Number 149423 by mathdanisur last updated on 05/Aug/21 $$\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}}…