Question Number 152052 by john_santu last updated on 25/Aug/21 $$\:\:\mathrm{knowns}\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{x}_{\mathrm{1}} }\:,\:\mathrm{x}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{x}_{\mathrm{2}} }\:,\:\mathrm{x}_{\mathrm{4}} =\frac{\mathrm{4}}{\mathrm{x}_{\mathrm{3}} } \\ $$$$,\:\mathrm{x}_{\mathrm{5}} =\frac{\mathrm{5}}{\mathrm{x}_{\mathrm{4}} }\:,\:…,\:\mathrm{x}_{\mathrm{8}} =\frac{\mathrm{8}}{\mathrm{x}_{\mathrm{7}} }.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{x}_{\mathrm{1}} ×\mathrm{x}_{\mathrm{2}}…
Question Number 152032 by mathdanisur last updated on 25/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 152034 by mathdanisur last updated on 25/Aug/21 $$\begin{cases}{\mathrm{3x}^{\mathrm{2}} \:-\:\mathrm{2y}\:=\:-\:\frac{\mathrm{17}}{\mathrm{3}}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{6x}\:=\:\mathrm{7}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{xy}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 25/Aug/21 $$\begin{cases}{\mathrm{3x}^{\mathrm{2}} \:-\:\mathrm{2y}\:=\:-\:\frac{\mathrm{17}}{\mathrm{3}}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{6x}\:=\:\mathrm{7}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{xy}\:=\:?\: \\…
Question Number 152030 by mathdanisur last updated on 25/Aug/21 Commented by mathdanisur last updated on 26/Aug/21 $$\boldsymbol{\mathrm{S}}\mathrm{er},\:\Omega_{{n}} +\Omega_{{n}+\mathrm{2}} +…+\Omega_{{n}×\mathrm{6}} =\mathrm{1} \\ $$ Commented by ghimisi…
Question Number 152029 by mathdanisur last updated on 25/Aug/21 Commented by ghimisi last updated on 25/Aug/21 $$\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{2}} \:\:{or}\:\:\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \\ $$ Commented by mathdanisur last updated…
Question Number 152019 by mathdanisur last updated on 25/Aug/21 $$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:{Li}_{\mathrm{2}} \left({x}\right)\:{log}\left(\mathrm{1}+{x}\right)\:{dx}\:=\:? \\ $$$${Li}_{\mathrm{2}} \left({x}\right)−{polylogaritm}\:{function} \\ $$ Answered by mnjuly1970 last updated on 25/Aug/21…
Question Number 20935 by Tinkutara last updated on 08/Sep/17 $$\mathrm{If}\:\mid{z}\:+\:\omega\mid^{\mathrm{2}} \:=\:\mid{z}\mid^{\mathrm{2}} \:+\:\mid\omega\mid^{\mathrm{2}} ,\:\mathrm{where}\:{z}\:\mathrm{and}\:\omega \\ $$$$\mathrm{are}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{real} \\ $$$$\left(\mathrm{2}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\left(\mathrm{3}\right)\:{z}\bar {\omega}\:+\:\bar {{z}}\omega\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{amp}\left(\frac{{z}}{\omega}\right)\:=\:\frac{\pi}{\mathrm{2}}…
Question Number 20933 by Tinkutara last updated on 08/Sep/17 $$\mathrm{If}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{satisfying} \\ $$$${z}\:+\:{z}^{−\mathrm{1}} \:=\:\mathrm{1},\:\mathrm{then}\:{z}^{{n}} \:+\:{z}^{−{n}} ,\:{n}\:\in\:{N},\:\mathrm{has} \\ $$$$\mathrm{the}\:\mathrm{value} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}\left(−\mathrm{1}\right)^{{n}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of} \\ $$$$\mathrm{3}…
Question Number 20934 by Tinkutara last updated on 08/Sep/17 $$\mathrm{If}\:{z}\:\mathrm{satisfies}\:\mid{z}\:−\:\mathrm{1}\mid\:<\:\mid{z}\:+\:\mathrm{3}\mid,\:\mathrm{then}\:\omega\:= \\ $$$$\mathrm{2}{z}\:+\:\mathrm{3}\:−\:{i}\:\mathrm{satisfies} \\ $$$$\left(\mathrm{1}\right)\:\mid\omega\:−\:\mathrm{5}\:−\:{i}\mid\:<\:\mid\omega\:+\:\mathrm{3}\:+\:{i}\mid \\ $$$$\left(\mathrm{2}\right)\:\mid\omega\:−\:\mathrm{5}\mid\:<\:\mid\omega\:+\:\mathrm{3}\mid \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Im}\:\left({i}\omega\right)\:>\:\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\mid\mathrm{arg}\left(\omega\:−\:\mathrm{1}\right)\mid\:<\:\frac{\pi}{\mathrm{2}} \\ $$ Commented by Tinkutara…
Question Number 20932 by Tinkutara last updated on 08/Sep/17 $$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and}\:{z}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that},\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \\ $$$$=\:\mathrm{1}\:\mathrm{and}\:{b}\:+\:{ic}\:=\:\left(\mathrm{1}\:+\:{a}\right){z},\:\mathrm{then}\:\frac{\mathrm{1}\:+\:{iz}}{\mathrm{1}\:−\:{iz}} \\ $$$$\mathrm{equals}. \\ $$$$\left(\mathrm{1}\right)\:\frac{{b}\:−\:{ic}}{\mathrm{1}\:−\:{ia}} \\ $$$$\left(\mathrm{2}\right)\:\frac{{a}\:+\:{ib}}{\mathrm{1}\:+\:{c}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}\:−\:{c}}{{a}\:−\:{ib}} \\…