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Question-198013

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-198014

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-197982

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-197989

Question Number 197989 by mr W last updated on 07/Oct/23 Commented by mr W last updated on 07/Oct/23 $$\left[\underline{{old}\:{question}\:{from}\:{ajfour}\:{sir}}\right] \\ $$$$\mathrm{3}\:{circles}\:{with}\:{radii}\:{p},\:{q},\:{r}\:{respectively} \\ $$$${touch}\:{each}\:{other}\:{as}\:{shown}.\:{find}\:{the}\: \\ $$$${maximum}\:{area}\:{of}\:\:{triangle}\:\Delta{ABC}…

Question-198018

Question Number 198018 by obia last updated on 07/Oct/23 Answered by a.lgnaoui last updated on 08/Oct/23 $$\boldsymbol{\mathrm{Z}}=\sqrt{\mathrm{3}}\:+\mathrm{1}+\:\boldsymbol{\mathrm{i}}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right) \\ $$$$ \\ $$$$\left.\bullet\mathrm{1}\right):\:\:\boldsymbol{\mathrm{Z}}^{\mathrm{2}} =\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\:−\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\:\right)+\mathrm{2}\boldsymbol{\mathrm{i}}\left(\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{3}}\:−\:\mathrm{1}\right) \\ $$$$\:\:\:=\mathrm{4}\sqrt{\mathrm{3}}\:\:+\mathrm{4}\boldsymbol{\mathrm{i}}\:\:=\mathrm{4}\left(\sqrt{\mathrm{3}}\:+\boldsymbol{\mathrm{i}}\right) \\…

Question-197967

Question Number 197967 by Blackpanther last updated on 06/Oct/23 Answered by mr W last updated on 06/Oct/23 $${f}\left({x}\right)={a}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{10}\right) \\ $$$$\frac{\mathrm{20}}{\mathrm{3}}={a}\left(\mathrm{0}+\mathrm{1}\right)\left(\mathrm{0}−\mathrm{10}\right) \\ $$$$\Rightarrow{a}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${at}\:{x}=\frac{−\mathrm{1}+\mathrm{10}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}}: \\…

determiner-le-total-de-nombres-de-5-chiffres-comprises-entre-10000-et-50000-divisibles-simultanement-par-5-et-9-sans-utiliser-les-formules-d-arrangement-et-de-combinaison-

Question Number 197960 by a.lgnaoui last updated on 06/Oct/23 $$\mathrm{determiner}\:\mathrm{le}\:\mathrm{total}\:\mathrm{de}\:\mathrm{nombres}\:\mathrm{de}\: \\ $$$$\mathrm{5}\:\mathrm{chiffres}\:\mathrm{comprises}\:\mathrm{entre}\:\mathrm{10000}\:\mathrm{et}\: \\ $$$$\mathrm{50000}\:\:\mathrm{divisibles}\:\mathrm{simultanement}\:\mathrm{par} \\ $$$$\mathrm{5}\:\mathrm{et}\:\mathrm{9}\:\:\: \\ $$$$\left(\mathrm{sans}\:\mathrm{utiliser}\:\mathrm{les}\:\mathrm{formules}\:\mathrm{d}\:\mathrm{arrangement}\right. \\ $$$$\left.\mathrm{et}\:\mathrm{de}\:\mathrm{combinaison}\right) \\ $$$$ \\ $$ Answered…