Question Number 150853 by mathdanisur last updated on 15/Aug/21 $$\mathrm{log}_{\mathrm{2021}} \:\sqrt{\mathrm{x}\::\:\sqrt{\mathrm{x}\::\:\sqrt{\mathrm{x}\::..}}}\:=\:\mathrm{674} \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$ Commented by liberty last updated on 16/Aug/21 $$\sqrt{\frac{\mathrm{x}}{\:\sqrt{\frac{\mathrm{x}}{\:\sqrt{\frac{\mathrm{x}}{\:\sqrt{\frac{\mathrm{x}}{\vdots}}}}}}}}\:=\:\mathrm{2021}^{\mathrm{674}} \\ $$$$\Leftrightarrow\sqrt{\frac{\mathrm{x}}{\mathrm{2021}^{\mathrm{674}}…
Question Number 85318 by Power last updated on 20/Mar/20 Commented by Power last updated on 21/Mar/20 $$+\mathrm{R}\left(\mathrm{x}\right)\:\:\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 150852 by mathdanisur last updated on 15/Aug/21 $$\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{3}\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{xyz}\:\leqslant\:\left(\frac{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:-\:\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 150840 by mathdanisur last updated on 15/Aug/21 $$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)}\:\mathrm{dx}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 17/Aug/21 $$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}}…
Question Number 150841 by mathdanisur last updated on 15/Aug/21 $$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{b}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 15/Aug/21…
Question Number 150838 by mathdanisur last updated on 15/Aug/21 Answered by amin96 last updated on 15/Aug/21 $$\frac{\left(\mathrm{3}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{5}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{7}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{4}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{6}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{8}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}\right)}=\frac{\left(\left(\mathrm{3}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{9}\right)\left(\left(\mathrm{5}^{\mathrm{2}}…
Question Number 150839 by mathdanisur last updated on 15/Aug/21 $$\mathrm{e}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{for}\:\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$ Answered by peter frank last updated on 15/Aug/21 $${e}^{{x}}…
Question Number 19741 by Tinkutara last updated on 15/Aug/17 $$\mathrm{If}\:{z}\:=\:{x}\:+\:{iy}\:\mathrm{and}\:\mathrm{arg}\left(\frac{{z}\:−\:\mathrm{2}}{{z}\:+\:\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{6}},\:\mathrm{then} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}. \\ $$ Commented by Tinkutara last updated on 15/Aug/17 $$\mathrm{I}\:\mathrm{found}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{equation}\:\mathrm{but}\:\mathrm{in}\:\mathrm{book} \\ $$$$“\mathrm{major}\:\mathrm{arc}''\:\mathrm{is}\:\mathrm{written}\:\mathrm{along}\:\mathrm{with}\:\mathrm{that} \\…
Question Number 19739 by Tinkutara last updated on 15/Aug/17 $$\mathrm{If}\:{z}\:=\:{x}\:+\:{iy}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number} \\ $$$$\mathrm{satisfying}\:\mid{z}\:+\:\frac{{i}}{\mathrm{2}}\mid^{\mathrm{2}} \:=\:\mid{z}\:−\:\frac{{i}}{\mathrm{2}}\mid^{\mathrm{2}} ,\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is} \\ $$ Answered by ajfour last updated on 15/Aug/17…
Question Number 19740 by Tinkutara last updated on 15/Aug/17 $$\mathrm{If}\:\mid{z}^{\mathrm{2}} \:−\:\mathrm{1}\mid\:=\:\mid{z}\mid^{\mathrm{2}} \:+\:\mathrm{1},\:\mathrm{then}\:{z}\:\mathrm{lies}\:\mathrm{on} \\ $$ Answered by ajfour last updated on 16/Aug/17 $$\:\mid\left(\mathrm{x}+\mathrm{iy}\right)^{\mathrm{2}} −\mathrm{1}\mid=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{1}…