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Question Number 186688 by norboyev last updated on 08/Feb/23 $${a}_{\mathrm{1}} =\mathrm{0} \\ $$$${a}_{\mathrm{2}} =\mathrm{1} \\ $$$${a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \\ $$$${a}_{\mathrm{885}} =? \\ $$ Commented by…

Question Number 186691 by 073 last updated on 08/Feb/23 Answered by Ar Brandon last updated on 08/Feb/23 $$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}{x}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)=\frac{\pi}{\mathrm{2}}−{x} \\ $$$$\int\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}{x}\right){dx}=\int\left(\frac{\pi}{\mathrm{2}}−{x}\right){dx} \\ $$…

Question Number 186680 by mathocean1 last updated on 08/Feb/23 $${study}\:{the}\:{convergence}\:{of}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{e}^{\left(\frac{\mathrm{1}}{{k}}\right)} \right) \\ $$ Terms of Service Privacy Policy Contact:…

Question Number 186675 by pascal889 last updated on 08/Feb/23 Answered by ARUNG_Brandon_MBU last updated on 08/Feb/23 $$\ast\mathrm{The}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{girls}\:\mathrm{to}\:\mathrm{boys}\:\mathrm{is}\:\mathrm{5}:\mathrm{6} \\ $$$$\Rightarrow\:\frac{{G}}{{B}}=\frac{\mathrm{5}}{\mathrm{6}}\:\Rightarrow{G}=\frac{\mathrm{5}}{\mathrm{6}}{B}\:…\mathrm{eqn}\left({i}\right) \\ $$$$\ast\mathrm{On}\:\mathrm{a}\:\mathrm{rainy}\:\mathrm{day}: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{6}}{G}\:\mathrm{were}\:\mathrm{absent}\:\Rightarrow\frac{\mathrm{5}}{\mathrm{6}}{G}\:\mathrm{were}\:\mathrm{present}. \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{B}\:\mathrm{were}\:\mathrm{absent}\:\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}{B}\:\mathrm{were}\:\mathrm{present}.…

Question Number 121111 by zakirullah last updated on 05/Nov/20 Commented by benjo_mathlover last updated on 05/Nov/20 $$\left(\mathrm{Q4}\right)\:\mathrm{n}\left(\mathrm{A}\cup\mathrm{B}\right)=\mathrm{50}+\mathrm{20}−\mathrm{10}=\mathrm{60} \\ $$ Commented by zakirullah last updated on…