Question Number 154613 by mathdanisur last updated on 20/Sep/21 $$\mathrm{There}\:\mathrm{are}\:\mathrm{3}\:\mathrm{or}\:\mathrm{more}\:\mathrm{profiles}\:\mathrm{of}\:\mathrm{this} \\ $$$$\mathrm{person}\:\mathrm{and}\:\mathrm{I}\:\mathrm{mentioned}\:\mathrm{it},\:\mathrm{he}\:\mathrm{knowes} \\ $$$$\mathrm{who}\:\mathrm{he}\:\mathrm{and}\:\mathrm{that}\:\mathrm{person}\:\mathrm{paints}\:\mathrm{what} \\ $$$$\mathrm{I}\:\mathrm{share}\:\mathrm{in}\:\mathrm{red},\:\mathrm{do}\:\mathrm{his}\:\mathrm{best},\:\mathrm{nothing}\:\mathrm{will} \\ $$$$\mathrm{chang}! \\ $$ Commented by liberty last updated…
Question Number 89070 by ajfour last updated on 15/Apr/20 $$\frac{\sqrt{{a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} }}\:=\:\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){x} \\…
Question Number 23537 by math solver last updated on 01/Nov/17 Commented by math solver last updated on 01/Nov/17 $$\mathrm{q}.\mathrm{7}\:? \\ $$ Commented by ajfour last…
Question Number 154596 by mathdanisur last updated on 19/Sep/21 $$\mathrm{let}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{z}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }}\:+\:\sqrt{\mathrm{2}}\:\geqslant\:\mathrm{2}\:\sqrt{\frac{\mathrm{z}}{\mathrm{x}+\mathrm{y}}}\:+\:\frac{\sqrt{\mathrm{2xy}}}{\mathrm{x}+\mathrm{y}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 89064 by john santu last updated on 15/Apr/20 $$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{…}}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{…}}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{…}}}}}}\:=\:… \\ $$ Answered by M±th+et£s last updated on 16/Apr/20 $$\mathrm{3}=\sqrt{\mathrm{1}+\mathrm{8}}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{16}}}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{25}}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{36}}}}}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{49}}}}}……} \\ $$$$=…………=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{1}+\mathrm{6}\sqrt{\mathrm{1}+…..}}}}}}…
Question Number 154599 by mathdanisur last updated on 19/Sep/21 Commented by mr W last updated on 21/Oct/21 $${see}\:{Q}\mathrm{157292} \\ $$ Answered by amumath last updated…
Question Number 89032 by M±th+et£s last updated on 14/Apr/20 $${find}\:{by}\:{using}\:{de}\:{moivre}'{s}\:{formula} \\ $$$${cos}\left(\mathrm{2}°\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 23479 by ANTARES_VY last updated on 31/Oct/17 $$\boldsymbol{\mathrm{Common}}\:\:\boldsymbol{\mathrm{solution}}. \\ $$$$\frac{\boldsymbol{\mathfrak{d}}}{\boldsymbol{\mathfrak{d}\mathrm{y}}}\left(\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{u}}\right)+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{u}}\right)=\mathrm{0}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 23471 by Tinkutara last updated on 31/Oct/17 $${Prove}\:{that} \\ $$$$\underset{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left(\frac{\mathrm{1}}{\:^{{n}} {C}_{{i}} }\:+\:\frac{\mathrm{1}}{\:^{{n}} {C}_{{j}} }\right)\:=\:\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{n}\:−\:{r}}{\:^{{n}} {C}_{{r}} }\:+\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{r}}{\:^{{n}} {C}_{{r}} }…
Question Number 154530 by mathdanisur last updated on 19/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com