Question Number 157478 by Ar Brandon last updated on 23/Oct/21 $$\mathrm{Sir}\:\mathrm{Tinku}-\mathrm{Tara}, \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{facing}\:\mathrm{some}\:\mathrm{difficulties}\:\mathrm{eversince}\:\mathrm{I}\:\mathrm{changed}\:\mathrm{my}\:\mathrm{device} \\ $$$$\mathrm{1}\bullet\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{save}\:\mathrm{my}\:\mathrm{work}\:\mathrm{as}\:\mathrm{images}\:\mathrm{to}\:\mathrm{my}\:\mathrm{new}\:\mathrm{device}. \\ $$$$\:\:\:\:\:\mathrm{In}\:\mathrm{case}\:\mathrm{I}\:\mathrm{have}\:\mathrm{to}\:\mathrm{upload}\:\mathrm{my}\:\mathrm{work}\:\mathrm{to}\:\mathrm{another}\:\mathrm{plateform}\:\mathrm{such}\:\mathrm{as}\:\mathrm{a}\: \\ $$$$\:\:\:\:\:\mathrm{whatsapp}\:\mathrm{group},\:\mathrm{I}'\mathrm{ll}\:\mathrm{just}\:\mathrm{have}\:\mathrm{to}\:\mathrm{send}\:\mathrm{it}\:\mathrm{directly}\:\mathrm{from}\:\mathrm{app}.\:\mathrm{Whereas} \\ $$$$\:\:\:\:\:\mathrm{with}\:\mathrm{my}\:\mathrm{previous}\:\mathrm{device}\:\mathrm{I}\:\mathrm{could}\:\mathrm{save}\:\mathrm{it}\:\mathrm{as}\:\mathrm{an}\:\mathrm{image}\:\mathrm{then}\:\mathrm{send}\:\mathrm{it}\:\mathrm{from} \\ $$$$\:\:\:\:\:\mathrm{gallery}\:\mathrm{while}\:\mathrm{on}\:\mathrm{another}\:\mathrm{plateforme}. \\ $$$$\mathrm{2}\bullet\:\mathrm{It}'\mathrm{s}\:\mathrm{impossible}\:\mathrm{for}\:\mathrm{me}\:\mathrm{to}\:\mathrm{send}\:\mathrm{images}\:\mathrm{after}\:\mathrm{inserting}\:\mathrm{page}\:\mathrm{breaks}.\:\mathrm{It}\:\mathrm{has}…
Question Number 91933 by liki last updated on 03/May/20 Commented by liki last updated on 03/May/20 $$…\mathrm{please}\:\mathrm{mr}\:\mathrm{w}\:\mathrm{help}\:\mathrm{me}\:,\mathrm{emergency}!< \\ $$ Commented by liki last updated on…
Question Number 157420 by Gbenga last updated on 23/Oct/21 $$\int\frac{\mathrm{1}}{\boldsymbol{{u}}^{\mathrm{4}} +\left(\mathrm{1}−\boldsymbol{{u}}\right)^{\mathrm{4}} }\boldsymbol{{du}} \\ $$ Answered by EvaNelle00 last updated on 23/Oct/21 $$\int\:\frac{{du}}{{u}^{\mathrm{4}} \left(\mathrm{1}\:+\:\left(\frac{\mathrm{1}−{u}}{{u}}\right)^{\mathrm{4}} \right)} \\…
Question Number 157388 by naka3546 last updated on 22/Oct/21 $${If}\:\:{m}\:\mathrm{tan}\:\left(\theta\:−\:\mathrm{30}°\right)\:=\:{n}\:\mathrm{tan}\:\left(\theta\:+\:\mathrm{12}°\right) \\ $$$$\mathrm{cos}\:\mathrm{2}\theta\:\:=\:? \\ $$ Commented by cortano last updated on 22/Oct/21 $$\:{m}\mathrm{tan}\:\left(\theta−\mathrm{30}°\right)={n}\mathrm{tan}\:\left(\mathrm{30}°+\theta−\mathrm{18}°\right) \\ $$$$\Rightarrow{m}\left(\frac{\mathrm{3tan}\:\theta−\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta}\right)={n}\left(\frac{\mathrm{tan}\:\left(\mathrm{30}°+\theta\right)−\mathrm{tan}\:\mathrm{18}°}{\mathrm{1}+\mathrm{tan}\:\left(\mathrm{30}°+\theta\right)\:\mathrm{tan}\:\mathrm{18}°}\right) \\…
Question Number 91837 by zainal tanjung last updated on 03/May/20 $$\left.\mathrm{1}\right).\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{5}}{\mathrm{n}+\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{n}+\mathrm{3}}\:\right)=… \\ $$$$ \\ $$$$\left.\mathrm{2}\right).\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4n}^{\mathrm{2}} −\mathrm{1}}\right)=… \\ $$$$ \\ $$$$\left.\mathrm{3}\right).\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty}…
Question Number 157374 by john_santu last updated on 22/Oct/21 $$ \\ $$I see there are idiot comments on this forum. a comment that tells someone…
Question Number 157341 by mathocean1 last updated on 22/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 157343 by mathocean1 last updated on 22/Oct/21 Answered by mindispower last updated on 23/Oct/21 $${X}^{\mathrm{2}} −{X}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{X}\in\left\{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$$${u}_{{n}} ={a}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{b}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\…
Question Number 91796 by zainal tanjung last updated on 03/May/20 $$\mathrm{Total}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{this}\:\mathrm{below}\:\mathrm{infinite}\:\mathrm{series}\:: \\ $$$$\left.\mathrm{1}\right).\:\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}+…\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}=… \\ $$$$\left.\mathrm{2}\right).\:\:\mathrm{0}.\mathrm{6}+\mathrm{0}.\mathrm{06}+\mathrm{0}.\mathrm{006}+…\frac{\mathrm{6}}{\mathrm{10}^{\mathrm{n}} }=… \\ $$ Commented by mathmax by abdo last updated…
Question Number 157329 by MathsFan last updated on 22/Oct/21 $$\:\boldsymbol{{whats}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\: \\ $$$$\:\:\:\:!\mathrm{5} \\ $$ Answered by puissant last updated on 24/Oct/21 $$!\mathrm{5}=\mathrm{5}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{120}}\right)=\mathrm{5}!\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{120}}\right) \\ $$$$=\:\mathrm{5}!\left(\frac{\mathrm{60}}{\mathrm{120}}−\frac{\mathrm{20}}{\mathrm{120}}+\frac{\mathrm{5}}{\mathrm{120}}−\frac{\mathrm{1}}{\mathrm{120}}\right)=\:\mathrm{120}×\frac{\mathrm{44}}{\mathrm{120}}\:=\:\mathrm{44}.. \\…