Question Number 216751 by sniper237 last updated on 17/Feb/25 $${Find}\:\:\:{Card}\left\{\left({A},{B},{C}\right)\in{P}\left({E}\right)^{\mathrm{3}} /\:{AUBUC}={E}\right\} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216746 by sniper237 last updated on 17/Feb/25 $${Calculate}\:\frac{\left(\mathrm{5}+{i}\right)^{\mathrm{4}} }{\mathrm{239}+{i}}\:{Then}\: \\ $$$${Prove}\:{the}\:{Machin}\:{formula} \\ $$$$\mathrm{4}{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{239}}\right)=\frac{\pi}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216742 by Tawa11 last updated on 17/Feb/25 $$\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{a}\:\:−\:\:\mathrm{1}\right)}{\mathrm{a}}\:\mathrm{da} \\ $$ Answered by sniper237 last updated on 17/Feb/25 $${Not}\:{defined}\:! \\ $$$${But}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 216737 by ahmed2025 last updated on 17/Feb/25 $$\:{find}\:{critical}\:{and}\:{local}\:{points}\:{for}\:{curve} \\ $$$$\:{y}={x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{2} \\ $$ Commented by MathematicalUser2357 last updated on 25/Feb/25 $$\mathrm{What}\:\mathrm{is}\:\mathrm{a}\:\mathrm{critical}\:\mathrm{point}?\:\mathrm{Critical}\:\mathrm{hit}\:\mathrm{point}? \\…
Question Number 216739 by ArshadS last updated on 17/Feb/25 $${Prove}\:{that}\:\:^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:−^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\:=\mathrm{1} \\ $$$${Question}#\mathrm{216694}\:{reposted}\:{for}\:{new}\:{answers} \\ $$ Answered by Rasheed.Sindhi last updated on 17/Feb/25 $$\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:=\mathrm{1} \\…
Question Number 216703 by LMKhit last updated on 16/Feb/25 Answered by issac last updated on 16/Feb/25 $$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{h}} }{{h}!\left(\frac{\mathrm{1}}{\mathrm{2}}−{h}\right)!}{x}^{\mathrm{2}{h}} \:,\:\mid{x}\mid\leq\mathrm{1} \\ $$$$\mathrm{because}\:{r}=\underset{{n}\rightarrow\infty} {\mathrm{lim}inf}\mid\:\frac{{a}_{{n}+\mathrm{1}}…
Question Number 216698 by sniper237 last updated on 16/Feb/25 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:^{\mathrm{3}} \sqrt{{x}+\mathrm{1}}\:−^{\mathrm{3}} \sqrt{{x}}\:\overset{?} {=}\:\mathrm{0} \\ $$ Answered by MrGaster last updated on 16/Feb/25 $$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}−\:^{\mathrm{3}} \sqrt{{x}}=\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 216694 by sniper237 last updated on 16/Feb/25 $${Prove}\:{that}\:\:^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:−^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\:=\mathrm{1} \\ $$ Answered by golsendro last updated on 16/Feb/25 $$\:\mathrm{let}\:\mathrm{x}=\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}+\mathrm{2}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$$\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}+\mathrm{2}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}−\mathrm{2}}−\mathrm{x}\:=\:\mathrm{0}\: \\…
Question Number 216695 by sniper237 last updated on 16/Feb/25 $$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{1}+{sinxcosx}}\overset{?} {=}\:\frac{\mathrm{4}\pi{ln}\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\: \\ $$ Answered by Ghisom last updated on 16/Feb/25 $$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}=\mathrm{8}\underset{\pi/\mathrm{4}}…
Question Number 216715 by Nadirhashim last updated on 16/Feb/25 $$\:\:\:\:\boldsymbol{{F}}{ind}\:\int\frac{\boldsymbol{{S}}{in}\left(\frac{\mathrm{5}{x}\:}{\mathrm{2}\:}\right)\:\:}{\boldsymbol{{S}}{in}\left(\frac{{x}\:}{\mathrm{2}\:}\right)\:\:\:\:\:}\:.\boldsymbol{{dx}}\:\:\: \\ $$ Answered by Frix last updated on 16/Feb/25 $$\int\frac{\mathrm{sin}\:\frac{\mathrm{5}{x}}{\mathrm{2}}}{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}}{dx}=\int\left(\mathrm{2cos}\:\mathrm{2}{x}\:+\mathrm{2cos}\:{x}\:+\mathrm{1}\right){dx}= \\ $$$$=\mathrm{sin}\:\mathrm{2}{x}\:+\mathrm{2sin}\:{x}\:+{x}+{C} \\ $$ Answered…