Question Number 206038 by BaliramKumar last updated on 05/Apr/24 $$\sqrt{\mathrm{1}\:+\:\mathrm{2023}\sqrt{\mathrm{1}\:+\:\mathrm{2024}\sqrt{\mathrm{1}+\:\mathrm{2025}\sqrt{\mathrm{1}\:+\:\mathrm{2026}\sqrt{\mathrm{1}\:+\:…………..\infty}}}}}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by mr W last updated on 05/Apr/24 $${x}+\mathrm{1}=\sqrt{\mathrm{1}+{x}\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)\sqrt{\mathrm{1}+…}}}} \\ $$$$\Rightarrow\mathrm{2024}=\sqrt{\mathrm{1}\:+\:\mathrm{2023}\sqrt{\mathrm{1}\:+\:\mathrm{2024}\sqrt{\mathrm{1}+\:\mathrm{2025}\sqrt{\mathrm{1}\:+\:\mathrm{2026}\sqrt{\mathrm{1}\:+\:…………..\infty}}}}} \\ $$…
Question Number 206024 by MaruMaru last updated on 05/Apr/24 $${proove} \\ $$$${e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$ Answered by Tinku Tara last updated on 05/Apr/24 $${e}^{{i}\pi} ={cos}\pi+{isin}\pi=−\mathrm{1}…
Question Number 206025 by cortano12 last updated on 05/Apr/24 $$\:\:\:\:\:\mathrm{2}^{\mathrm{2024}} \:=\:{x}\:\left({mod}\:\mathrm{10}\right)\: \\ $$ Answered by BaliramKumar last updated on 05/Apr/24 $$\mathrm{Find}\:\mathrm{unit}\:\mathrm{digit} \\ $$$$\mathrm{2}^{\mathrm{2024}} \:=\:\mathrm{2}^{\mathrm{4k}\:+\:\mathrm{4}} \:=\:\mathrm{2}^{\mathrm{4}}…
Question Number 206020 by NANIGOPAL last updated on 05/Apr/24 Answered by cortano12 last updated on 05/Apr/24 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}\:−\mathrm{2tan}\:{x}\right)}{\mathrm{4sin}\:^{\mathrm{4}} {x}} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}\mathrm{tan}\:{x}\left(\mathrm{1}−\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}\right)\right)}{\mathrm{4sin}\:^{\mathrm{4}} {x}}…
Question Number 206053 by mr W last updated on 05/Apr/24 Commented by mr W last updated on 05/Apr/24 $${an}\:{unsolved}\:{old}\:{question}\:\left({Q}\mathrm{182165}\right) \\ $$ Answered by mr W…
Question Number 206048 by MATHEMATICSAM last updated on 05/Apr/24 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{2}^{\mathrm{sin}^{\mathrm{2}} \theta} \:+\:\mathrm{2}^{\mathrm{cos}^{\mathrm{2}} \theta} \:\geqslant\:\mathrm{2}\sqrt{\mathrm{2}}. \\ $$ Answered by A5T last updated on 05/Apr/24 $$\mathrm{2}^{{sin}^{\mathrm{2}} \theta}…
Question Number 205993 by otchereabdullai@gmail.com last updated on 04/Apr/24 $$\:{Given}\:{that}\:\left(\mathrm{3}−\sqrt{{n}}\right)^{\mathrm{2}} ={m}−\mathrm{6}\sqrt{\mathrm{2}}\:{where} \\ $$$${m},{n}\:{are}\:{positive}\:{integers}\:{find}\:{m}−{n} \\ $$ Answered by A5T last updated on 04/Apr/24 $$\mathrm{9}+{n}−\mathrm{6}\sqrt{{n}}={m}−\mathrm{6}\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{9}+{n}={m}\:\wedge\:{n}=\mathrm{2}\Rightarrow{m}−{n}=\mathrm{9} \\ $$…
Question Number 205988 by cortano12 last updated on 04/Apr/24 $$\:\:{Given}\:{f}\left({x}+\mathrm{1}\right)=\mathrm{2}^{{f}\left({x}\right)} .{f}\left(\mathrm{1}\right) \\ $$$$\:\:{and}\:{f}\left(\mathrm{1}\right)=\:\mathrm{16}\: \\ $$$$\:\:{then}\:{f}\left(\mathrm{2016}\right)=? \\ $$ Answered by Berbere last updated on 04/Apr/24 $${f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{20}}…
Question Number 205990 by MATHEMATICSAM last updated on 04/Apr/24 $$\mathrm{If}\:{x}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{then}\:\frac{\sqrt{\mathrm{1}\:+\:{x}}\:+\:\sqrt{\mathrm{1}\:−\:{x}}}{\:\sqrt{\mathrm{1}\:+\:{x}}\:−\:\sqrt{\mathrm{1}\:−\:{x}}}\:=\:? \\ $$ Answered by cortano12 last updated on 04/Apr/24 $$\:\frac{\left(\sqrt{\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}\:\right)^{\mathrm{2}} }{\mathrm{2}{x}}\:=\:\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{2}{x}}\: \\ $$$$\:=\:\frac{\mathrm{1}}{{x}}\:+\:\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}}\:…
Question Number 206007 by SANOGO last updated on 04/Apr/24 $$\left({E},<,>\:\right):\:\:\:{prouve} \\ $$$$<{x},{y}>=\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}={o}} {\overset{\mathrm{3}} {\sum}}{i}^{{k}} \mid\mid{x}\:+\:{i}^{{k}} {y}\mid\mid^{\mathrm{2}} \\ $$ Commented by Berbere last updated on 04/Apr/24…