Question Number 128035 by rs4089 last updated on 03/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128030 by Agnibhoo last updated on 03/Jan/21 $$\:\mathrm{99}\:×\:\mathrm{99}\:=\:\mathrm{9801} \\ $$$$\:\mathrm{999}\:×\:\mathrm{999}\:=\:\mathrm{998001} \\ $$$$\:\mathrm{9999}\:×\:\mathrm{9999}\:=\:\mathrm{99980001} \\ $$$$\:\mathrm{99999}\:×\:\mathrm{99999}\:=\:? \\ $$$$\:\mathrm{999999}\:×\:\mathrm{999999}\:=\:? \\ $$ Answered by Geovanek last updated…
Question Number 62494 by Tawa1 last updated on 21/Jun/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128029 by Algoritm last updated on 03/Jan/21 Answered by MJS_new last updated on 03/Jan/21 $$\mathrm{3}^{−\mathrm{5}/\mathrm{4}} <{x}<\mathrm{3}^{−\mathrm{1}} \vee\mathrm{3}^{\mathrm{5}} <{x} \\ $$$$\mathrm{let}\:{x}=\mathrm{3}^{{t}} \:\mathrm{to}\:\mathrm{get}\:\mathrm{this}\:\mathrm{very}\:\mathrm{easily} \\ $$…
Question Number 128026 by Ar Brandon last updated on 03/Jan/21 $$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{2}\pi}{\mathrm{3}}} \frac{\left(\mathrm{arcsinhx}\right)^{\mathrm{2}} }{\mathrm{2cosx}}\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62489 by Tawa1 last updated on 21/Jun/19 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:+\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}{\:\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:−\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}\:\:=\:\:\mathrm{3} \\ $$ Answered by som(math1967) last updated on 22/Jun/19 $$\frac{\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}+\sqrt{\mathrm{2}−{x}}−\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}−\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{3}+\mathrm{1}}{\mathrm{3}−\mathrm{1}}\:\:\bigstar \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}−{x}}}{\mathrm{2}\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{4}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}=\mathrm{4}\:\:\bigstar\bigstar \\…
Question Number 128025 by Algoritm last updated on 03/Jan/21 Answered by Olaf last updated on 03/Jan/21 $$\mathrm{Let}\:\Phi_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {e}^{\mathrm{cos}\theta} \mathrm{cos}\left({n}\theta−\mathrm{sin}\theta\right){d}\theta \\ $$$$\mathrm{Let}\:\Omega_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi}…
Question Number 128023 by mnjuly1970 last updated on 04/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:\:{calculus}…= \\ $$$$\:\:{Titu}'{s}\:{lemma}:: \\ $$$$\:{for}\:{any}\:{positive}\:{numbers}\:: \\ $$$${a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,…,{a}_{{n}} \:,\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,…,{b}_{{n}} \\ $$$$\:{we}\:{have}: \\ $$$$\:\frac{\left({a}_{\mathrm{1}}…
Question Number 62486 by ajfour last updated on 21/Jun/19 Commented by ajfour last updated on 21/Jun/19 $${Find}\:{maximum}\:{perimeter}\:{of} \\ $$$${quadrilateral}\:{ABCD}. \\ $$ Answered by mr W…
Question Number 128015 by Algoritm last updated on 03/Jan/21 Answered by MJS_new last updated on 03/Jan/21 $${x}=\sqrt[{\mathrm{8}}]{\mathrm{2}} \\ $$ Answered by MJS_new last updated on…