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 209232 by alcohol last updated on 04/Jul/24 $${u}_{\mathrm{0}} \:=\:{a},\:{u}_{{n}+\mathrm{1}} \:=\:\sqrt{{u}_{{n}} {v}_{{n}} } \\ $$$$\left.{v}_{\mathrm{0}} \:=\:{b}\:\in\:\right]\mathrm{0},\mathrm{1}\left[\:,\:{v}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)}\right. \\ $$$$\bullet\:{show}\:{that}\:{a}\leqslant{u}_{{n}} \leqslant{u}_{{n}+\mathrm{1}} \leqslant{v}_{{n}} \leqslant{v}_{{n}+\mathrm{1}}…
Question Number 209217 by mnjuly1970 last updated on 04/Jul/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{calculate}}\:: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}=\:\int_{\mathrm{0}\:} ^{\:\infty} \frac{\:{tan}^{\:−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}\:+\:{x}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\: \\ $$$$ \\ $$…
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 $${Same}\:{arrangement}: \\ $$$$\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}\left(>\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}\right)>\sqrt{\mathrm{7}}−\sqrt{\mathrm{5}}>\sqrt{\mathrm{13}}−\sqrt{\mathrm{11}}>\sqrt{\mathrm{19}}−\sqrt{\mathrm{17}} \\…
Question Number 209193 by Tawa11 last updated on 03/Jul/24 A pin 6cm high is placed in front of a diverging lens of focal length 15cm,…
Question Number 209211 by SonGoku last updated on 03/Jul/24 Commented by A5T last updated on 03/Jul/24 $${AC}=\sqrt{{AB}^{\mathrm{2}} +{BC}^{\mathrm{2}} −\mathrm{2}{AB}×{BCcos}\mathrm{123}}\approx\mathrm{74}.\mathrm{738} \\ $$ Terms of Service Privacy…
Question Number 209206 by Frix last updated on 03/Jul/24 $$\mathrm{2}\:\mathrm{YouTube}\:\mathrm{channels}\:\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{might} \\ $$$$\mathrm{find}\:\mathrm{useful}. \\ $$ Commented by Frix last updated on 03/Jul/24 https://youtube.com/@primenewtons?si=lMR_ztxUBp9VG6GH Commented by Frix…
Question Number 209185 by Shivamsharma last updated on 03/Jul/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 209187 by mnjuly1970 last updated on 03/Jul/24 $$ \\ $$$$\:\:\:::\:\:\:\alpha\:,\:\beta\:\:{and}\:\:\gamma\:\:{are}\:{roots}\:{of}\:{the} \\ $$$$\:\:\:\:\:{following}\:\:{equation}\:.\:{Find}\:{the} \\ $$$$\:\:\:\:\:{value}\:\:{of}\:\:\:''\:\:\mathrm{F}\:\:''\::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{E}{quation}\::\:\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{2}{x}\:\:−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{F}\::=\:\alpha^{\:\mathrm{5}} \:+\:\beta^{\:\mathrm{5}} \:+\:\gamma^{\:\mathrm{5}}…
Question Number 209116 by alcohol last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $${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)…