Question Number 30257 by mondodotto@gmail.com last updated on 19/Feb/18 Commented by rahul 19 last updated on 19/Feb/18 $$\mathrm{Sandwhich}\:\mathrm{theorem}\:! \\ $$ Commented by abdo imad last…
Question Number 30258 by mondodotto@gmail.com last updated on 19/Feb/18 Answered by $@ty@m last updated on 19/Feb/18 $${Let}\:\mathrm{sec}\:{x}−\mathrm{1}={z} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} {xdx}={dz} \\ $$$${dx}=\frac{{dz}}{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}}=\frac{{dz}}{{z}\left({z}+\mathrm{2}\right)} \\ $$$${I}=\int\frac{{dx}}{{z}^{\mathrm{2}}…
Question Number 30256 by mondodotto@gmail.com last updated on 19/Feb/18 Commented by abdo imad last updated on 19/Feb/18 $${in}\:{this}\:{case}\:{is}\:{better}\:{to}\:{use}\:{hospital}\:{theorem}\:{let}\:{put} \\ $$$${u}\left({x}\right)={x}^{{x}} \:−{x}\:{and}\:{v}\left({x}\right)=\mathrm{1}−{x}\:+{lnx}\:{wehave}\:{u}\left(\mathrm{1}\right)={v}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${u}\left({x}\right)={e}^{{xlnx}} −{x}\Rightarrow{u}^{'} \left({x}\right)=\left(\mathrm{1}+{lnx}\right){e}^{{xlnx}}…
Question Number 95782 by john santu last updated on 27/May/20 $$\mathrm{it}\:\mathrm{looks}\:\mathrm{version}\:\mathrm{2}.\mathrm{077}\:\mathrm{has}\:\mathrm{a}\: \\ $$$$\mathrm{problem}.\:\:\mathrm{for}\:\mathrm{very}\:\mathrm{long}\:\mathrm{loading} \\ $$ Commented by Tinku Tara last updated on 27/May/20 $$\mathrm{Initial}\:\mathrm{loading}?\:\mathrm{Immediately} \\…
Question Number 161322 by kapoorshah last updated on 16/Dec/21 Commented by cortano last updated on 16/Dec/21 $$\mathrm{sin}\:\left(\mathrm{30}°+\mathrm{10}°+\alpha\right)={b} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{10}°+\alpha\right)+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{10}°+\alpha\right)={b} \\ $$$$\mathrm{cos}\:\left(\mathrm{10}°+\alpha\right)+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{10}°+\alpha\right)=\mathrm{2}{b} \\ $$$$\sqrt{\mathrm{3}}\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {t}}\:=\:\mathrm{2}{b}−\mathrm{cos}\:{t}\:;\:{t}=\mathrm{10}°+\alpha \\…
Question Number 161316 by otchereabdullai@gmail.com last updated on 16/Dec/21 $$\mathrm{Three}\:\mathrm{quarters}\:\mathrm{of}\:\mathrm{a}\:\mathrm{number}\:\mathrm{added}\:\mathrm{to} \\ $$$$\mathrm{two}\:\mathrm{and}\:\mathrm{a}\:\mathrm{half}\:\mathrm{of}\:\mathrm{that}\:\mathrm{number}\:\mathrm{gives}\: \\ $$$$\mathrm{13}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{number} \\ $$$$ \\ $$ Answered by FelipeLz last updated on 16/Dec/21…
Question Number 30245 by .none. last updated on 19/Feb/18 $$\mathrm{1}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} +\bullet\bullet\bullet+\mathrm{17}^{\mathrm{2}} −\mathrm{19}^{\mathrm{2}} \\ $$ Commented by Penguin last updated on 19/Feb/18 $$=\underset{{n}=\mathrm{1}}…
Question Number 95773 by Fikret last updated on 27/May/20 Answered by Fikret last updated on 27/May/20 $$ \\ $$$${find}\:{the}\:{geometrical}\:{centers}\:{of}\: \\ $$$${the}\:{following}\:{sections} \\ $$ Answered by…
Question Number 95770 by MJS last updated on 27/May/20 $$\mathrm{the}\:\mathrm{app}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{work}\:\mathrm{properly}\:\mathrm{for}\:\mathrm{me}\:\mathrm{any} \\ $$$$\mathrm{more}.\:\mathrm{I}\:\mathrm{open}\:\mathrm{it}\:\mathrm{and}\:\mathrm{most}\:\mathrm{of}\:\mathrm{the}\:\mathrm{time}\:\mathrm{the} \\ $$$$\mathrm{home}\:\mathrm{screen}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{show}\:\mathrm{the}\:\mathrm{forum}. \\ $$$$\mathrm{refreshing}\:\mathrm{or}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{switch}\:\mathrm{to}\:\mathrm{the}\:\mathrm{forum} \\ $$$$\mathrm{ends}\:\mathrm{up}\:\mathrm{in}\:\mathrm{an}\:\mathrm{endless}\:\mathrm{turning}\:\mathrm{loop}… \\ $$ Commented by PRITHWISH SEN 2…
Question Number 161294 by floor(10²Eta[1]) last updated on 15/Dec/21 $$\int_{−\mathrm{2}} ^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{3}} \mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$ Answered by puissant last updated on 20/Dec/21 $${f}\left({x}\right)={x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}}…