Question Number 208482 by efronzo1 last updated on 17/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208515 by mokys last updated on 17/Jun/24 $$\int_{\mathrm{0}} ^{\:\infty} \:\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208445 by alcohol last updated on 16/Jun/24 $$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} ,\:{u}_{\mathrm{0}} \in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} }\:=\:{f}\left({u}_{{n}} ^{\mathrm{2}} \right)\:;\:{f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }…
Question Number 208447 by efronzo1 last updated on 16/Jun/24 $$\:\:\underbrace{\:} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208440 by liuxinnan last updated on 16/Jun/24 $${a}_{\mathrm{1}} >{a}_{\mathrm{2}} >{a}_{\mathrm{3}} >…>{a}_{{n}} >\mathrm{0} \\ $$$${b}_{\mathrm{1}} >{b}_{\mathrm{2}} >{b}_{\mathrm{3}} >…>{b}_{{n}} >\mathrm{0} \\ $$$${prove} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}}…
Question Number 208441 by efronzo1 last updated on 16/Jun/24 Answered by Etimbuk last updated on 16/Jun/24 $$\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{3}\:} =\:\mathrm{y}^{\mathrm{3}} \:\Rightarrow\:\mathrm{x}\:=\:\mathrm{y} \\ $$$$…
Question Number 208468 by Tawa11 last updated on 16/Jun/24 Answered by A5T last updated on 17/Jun/24 $$\frac{{OD}×{OC}}{{AO}×{OB}}=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\mathrm{9}}{\mathrm{25}}\Rightarrow{k}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\frac{{AB}}{{DC}}=\frac{\mathrm{5}}{\mathrm{3}};\:{AB}=\mathrm{5}{x},{DC}=\mathrm{3}{x} \\ $$$$\left[{ABCD}\right]=\frac{\left(\mathrm{8}{x}\right){h}}{\mathrm{2}}=\mathrm{4}{xh} \\ $$$$\left[{OBC}\right]=\left[{OAD}\right]=\frac{\mathrm{5}{xh}}{\mathrm{2}}−\mathrm{25}=\frac{\mathrm{3}{xh}}{\mathrm{2}}−\mathrm{9}\Rightarrow{xh}=\mathrm{16} \\…
Question Number 208469 by hardmath last updated on 16/Jun/24 $$\mathrm{Find}:\:\:\:\:\:\frac{\mathrm{61}^{\mathrm{3}} \:\:+\:\:\mathrm{24}^{\mathrm{3}} }{\mathrm{61}^{\mathrm{3}} \:\:+\:\:\mathrm{37}^{\mathrm{3}} }\:\:=\:\:? \\ $$ Answered by Frix last updated on 16/Jun/24 $${a}=\mathrm{61};\:{b}=\mathrm{24};\:{a}−{b}=\mathrm{37} \\…
Question Number 208437 by efronzo1 last updated on 16/Jun/24 Answered by som(math1967) last updated on 16/Jun/24 $$\:{a}={sin}\mathrm{84}+{cos}\mathrm{66} \\ $$$$\Rightarrow{a}={sin}\mathrm{84}+{sin}\mathrm{24} \\ $$$$\Rightarrow{a}=\mathrm{2}{sin}\mathrm{54}{cos}\mathrm{30} \\ $$$$\:\therefore{a}=\mathrm{2}×\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{4}} \\ $$$${b}={sin}\mathrm{48}−{sin}\mathrm{12}…
Question Number 208465 by Tawa11 last updated on 16/Jun/24 Answered by A5T last updated on 17/Jun/24 $$\frac{\mathrm{3}\sqrt{\mathrm{2}}}{{s}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}}\Rightarrow{s}=\frac{\mathrm{10}\sqrt{\mathrm{10}}}{\mathrm{5}}=\mathrm{2}\sqrt{\mathrm{10}}\Rightarrow{s}^{\mathrm{2}} =\mathrm{40} \\ $$ Commented by Tawa11 last updated…