Question Number 198031 by cortano12 last updated on 08/Oct/23 Answered by AST last updated on 08/Oct/23 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{20}{ab}\Rightarrow\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{18}{ab}…\left({i}\right) \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{19}{ab}\Rightarrow\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}}…
Question Number 198025 by mr W last updated on 08/Oct/23 Commented by mr W last updated on 08/Oct/23 $${prove}\:{PQ}={RS} \\ $$ Answered by mahdipoor last…
Question Number 198022 by cortano12 last updated on 08/Oct/23 $$\:\:\mathrm{Five}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{selected}\:\mathrm{from} \\ $$ Answered by mr W last updated on 08/Oct/23 $${OOOXX}\:\:\Rightarrow\mathrm{2} \\ $$$${OOOXY}\:\:\Rightarrow\mathrm{2}×{C}_{\mathrm{2}} ^{\mathrm{4}} =\mathrm{12}…
Question Number 198023 by riyana last updated on 08/Oct/23 Answered by Rasheed.Sindhi last updated on 08/Oct/23 $$\frac{\:\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{4}}{\mathrm{7}}\:}{\:\mathrm{1}\frac{\mathrm{5}}{\mathrm{7}}+\frac{\mathrm{9}}{\mathrm{14}}\:}\centerdot\frac{\mathrm{11}}{\mathrm{26}} \\ $$$$\frac{\:\frac{\mathrm{21}−\mathrm{8}}{\mathrm{14}}\:}{\:\frac{\mathrm{12}}{\mathrm{7}}+\frac{\mathrm{9}}{\mathrm{14}}\:}\:\centerdot\:\frac{\mathrm{11}}{\mathrm{26}} \\ $$$$\frac{\:\frac{\mathrm{13}}{\mathrm{14}}\:}{\:\frac{\mathrm{24}+\mathrm{9}}{\mathrm{14}}\:}\:\centerdot\:\frac{\mathrm{11}}{\mathrm{26}} \\ $$$$\frac{\cancel{\overset{\mathrm{1}} {\mathrm{13}}}}{\cancel{\underset{\mathrm{3}} {\mathrm{33}}}}\:\centerdot\:\frac{\cancel{\overset{\mathrm{1}}…
Question Number 197980 by mr W last updated on 07/Oct/23 Commented by mr W last updated on 07/Oct/23 $${please}\:{exact}\:{solution}! \\ $$ Commented by Blackpanther last…
Question Number 198013 by sonukgindia last updated on 07/Oct/23 Answered by witcher3 last updated on 10/Oct/23 $$\mathrm{nice}\:\mathrm{one}\:\:\mathrm{sir} \\ $$$$\mathrm{are}\:\mathrm{you}\:\mathrm{sur}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression} \\ $$ Terms of Service Privacy…
Question Number 198014 by hardmath last updated on 07/Oct/23 Answered by mahdipoor last updated on 07/Oct/23 $$=\underset{{x}=−\mathrm{1}} {\overset{−\infty} {\sum}}\left(\underset{{y}={x}−\mathrm{1}} {\overset{−\infty} {\sum}}\mathrm{2}^{{x}+\mathrm{1}} ×\mathrm{3}^{{y}} ×\mathrm{5}\right)=\mathrm{5}\left(\underset{{x}=−\mathrm{1}} {\overset{−\infty} {\sum}}\mathrm{2}^{{x}+\mathrm{1}}…
Question Number 197982 by hardmath last updated on 07/Oct/23 Answered by witcher3 last updated on 11/Oct/23 $$\mathrm{sin}\left(\frac{\mathrm{12}\pi}{\mathrm{180}}\right)\mathrm{sin}\left(\frac{\mathrm{15}\pi}{\mathrm{180}}\right)\mathrm{sin}\left(\frac{\mathrm{51}\pi}{\mathrm{180}}\right)\mathrm{sin}\left(\frac{\mathrm{57}\pi}{\mathrm{180}}\right)\mathrm{sin}\left(\frac{\mathrm{63}\pi}{\mathrm{180}}\right) \\ $$$$\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{60}}\right)\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{60}}\right)\mathrm{sin}\left(\frac{\mathrm{17}\pi}{\mathrm{60}}\right)\mathrm{sin}\left(\frac{\mathrm{19}\pi}{\mathrm{60}}\right)\mathrm{sin}\left(\frac{\mathrm{21}\pi}{\mathrm{60}}\right) \\ $$$$\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{60}}\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{30}}\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{60}}\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{12}}\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{sin}\left(\frac{\mathrm{17}\pi}{\mathrm{60}}\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}+\frac{\pi}{\mathrm{12}}\right)…
Question Number 198001 by mathlove last updated on 07/Oct/23 $$\int\left({e}\right)^{\left({x}\right)^{{lnx}} } \:{dx}=? \\ $$ Commented by Frix last updated on 07/Oct/23 $$\left(\mathrm{e}\right)^{\left({x}\right)^{\mathrm{ln}\:{x}} } =\mathrm{e}^{\left({x}^{\mathrm{ln}\:{x}} \right)}…
Question Number 197998 by mathlove last updated on 07/Oct/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} }=? \\ $$$${with}\:{out}\:{L}'{Hospital}\:{rul} \\ $$ Answered by witcher3 last updated on 07/Oct/23 $$\mathrm{x}=\mathrm{sh}\left(\mathrm{t}\right),\mathrm{t}\rightarrow\mathrm{0}…