Question Number 206063 by MATHEMATICSAM last updated on 06/Apr/24 $$\mathrm{If}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\left({y}\:+\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{2}} }\right)\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{find}\:\left({x}\:+\:{y}\right)^{\mathrm{2}} . \\ $$ Answered by mr W last updated on 06/Apr/24…
Question Number 206082 by hardmath last updated on 06/Apr/24 $$\mathrm{If}\:\:\:\mathrm{cos}\boldsymbol{\alpha}\:=\:\mathrm{sin}\boldsymbol{\alpha}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Find}\:\:\:\mathrm{sin2}\boldsymbol{\alpha}\:=\:? \\ $$ Answered by MATHEMATICSAM last updated on 29/Apr/24 $$\mathrm{cos}\alpha\:=\:\mathrm{sin}\alpha\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\mathrm{cos}\alpha\:−\:\mathrm{sin}\alpha\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\…
Question Number 206025 by cortano12 last updated on 05/Apr/24 $$\:\:\:\:\:\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} \\ $$$$\mathrm{2}^{\mathrm{2024}} \:=\:\mathrm{2}^{\mathrm{4k}\:+\:\mathrm{4}} \:=\:\mathrm{2}^{\mathrm{4}}…
Question Number 205988 by cortano12 last updated on 04/Apr/24 $$\:\:{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}}\: \\ $$$$\:=\:\frac{\mathrm{1}}{{x}}\:+\:\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}}\:…
Question Number 205941 by cortano12 last updated on 03/Apr/24 $$\:\:\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}−\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}−\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{3}}}}}}\:=?\: \\ $$$$\: \\ $$ Answered by mr W last updated on 03/Apr/24 $$\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}−\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}−\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{3}}}}}} \\ $$$$=\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}−\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}−\frac{\mathrm{3}}{\mathrm{4}}}}}…
Question Number 205971 by mnjuly1970 last updated on 03/Apr/24 $$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{9}\:\:\mathscr{M}{athematical}\:\:\:\:\mathscr{A}{nalysis}\:\left(\:{I}\:\right) \\ $$$$\:\:\left({X}\:,\:{d}\:\right)\:{is}\:{a}\:{metric}\:{space}\:{and}\:\:\:\:\:\:\: \\ $$$$\:\:\:\left\{\:{p}_{{n}} \right\}_{{n}=\mathrm{1}} ^{\infty} {is}\:{a}\:{sequence}\:{in}\:{X}\: \\ $$$$\:\:\:\:\:{such}\:{that}\:,\:{p}_{{n}} \overset{{convergent}} {\rightarrow}\:{p}\:.\:{If}\:,\:{K}=\:\left\{{p}_{{n}} \right\}_{{n}=\mathrm{1}} ^{\infty}…
Question Number 205914 by mr W last updated on 02/Apr/24 $${if}\:{a}+{b}+{c}+{d}+{e}+{f}=\mathrm{10}\:{and} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} =\mathrm{25},\:{find} \\ $$$${a}_{{min}} \:{and}\:{f}_{{max}} . \\ $$…
Question Number 205884 by Abdullahrussell last updated on 01/Apr/24 Commented by Frix last updated on 01/Apr/24 $${a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 205885 by hardmath last updated on 01/Apr/24 $$\mathrm{Find}:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{}\begin{pmatrix}{\mathrm{2n}}\\{\:\:\mathrm{n}}\end{pmatrix}\:=\:? \\ $$ Answered by MM42 last updated on 03/Apr/24 $$\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}=\frac{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)…\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\mathrm{3}×\mathrm{2}×\mathrm{1}} \\ $$$${a}=\sqrt[{{n}}]{}\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix} \\ $$$$\Rightarrow{lna}=\frac{\mathrm{1}}{{n}}\left[{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{2}}{{n}−\mathrm{2}}\right)+…+{ln}\left(\mathrm{2}+\frac{{n}−\mathrm{3}}{\mathrm{3}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{2}}{\mathrm{2}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{1}}{\mathrm{1}}\right)\right.…