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Question-211414

Question Number 211414 by AlagaIbile last updated on 08/Sep/24 Answered by Frix last updated on 08/Sep/24 $$\mathrm{Let}\:{x}_{{j}} \leqslant{x}_{{j}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{x}_{\mathrm{1}} }=\mathrm{1}\:\Rightarrow\:\mathrm{max}\:\left({x}_{\mathrm{1}} \right)\:=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}_{\mathrm{1}} }+\frac{\mathrm{1}}{{x}_{\mathrm{2}}…

F-0-0-F-1-1-F-n-1-F-n-F-n-1-prove-1-89-i-1-10-i-F-i-1-

Question Number 211370 by liuxinnan last updated on 07/Sep/24 $${F}\left(\mathrm{0}\right)=\mathrm{0}\:\:\:\:\:\:\:{F}\left(\mathrm{1}\right)=\mathrm{1}\:\:\:\:{F}\left({n}+\mathrm{1}\right)={F}\left({n}\right)+{F}\left({n}−\mathrm{1}\right) \\ $$$${prove}: \\ $$$$\frac{\mathrm{1}}{\mathrm{89}}=\underset{{i}=\mathrm{1}} {\overset{+\infty} {\sum}}\mathrm{10}^{−{i}} {F}\left({i}−\mathrm{1}\right) \\ $$ Answered by mr W last updated…

Question-211258

Question Number 211258 by otchereabdullai@gmail.com last updated on 02/Sep/24 Answered by A5T last updated on 02/Sep/24 $$\Rightarrow{tan}\mathrm{2}\theta=\frac{\mathrm{28}}{\mathrm{9}}\Rightarrow\theta=\frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{28}}{\mathrm{9}}\right)}{\mathrm{2}}\approx\mathrm{36}.\mathrm{091}° \\ $$ Terms of Service Privacy Policy…