Question Number 163168 by kdaramaths last updated on 04/Jan/22 Commented by mr W last updated on 04/Jan/22 $${question}\:{is}\:{wrong}! \\ $$$${there}\:{is}\:{no}\:{solution}!\:{see}\:{my}\:{proof} \\ $$$${below}. \\ $$ Answered…
Question Number 97576 by Rio Michael last updated on 08/Jun/20 $$\:\mathrm{Show}\:\mathrm{that}\:{RE}\left[\frac{\mathrm{1}}{\mathrm{1}−{z}}\right]=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{where}\:{z}\:=\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\theta \\ $$$$ \\ $$ Answered by smridha last updated on 08/Jun/20 $$\boldsymbol{{RE}}\left[\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{{e}}^{\boldsymbol{{i}\theta}} }\right]=\boldsymbol{{RE}}\left[\frac{\mathrm{1}+{e}^{{i}\boldsymbol{\theta}} }{\mathrm{1}−{e}^{\mathrm{2}{i}\boldsymbol{\theta}}…
Question Number 97413 by Rio Michael last updated on 08/Jun/20 $$\mathrm{Given}\:\mathrm{that}\:\omega\:=\:{e}^{{i}\theta} ,\:\theta\neq\:{n}\pi\:,\:{n}\:\in\mathbb{N} \\ $$$$\mathrm{show}\:\mathrm{that}\:\left(\mathrm{1}\:+\:\omega\right)^{{n}} \:=\:\mathrm{2}^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}\left({in}\theta\right)} \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{out}\:\mathrm{on}\:\mathrm{this},\:\mathrm{i}'\mathrm{ve}\:\mathrm{stumbled}\:\mathrm{on}\:\mathrm{it}. \\ $$ Answered by mathmax by abdo…
Question Number 31719 by Penguin last updated on 13/Mar/18 Commented by Penguin last updated on 13/Mar/18 Commented by Penguin last updated on 13/Mar/18 Commented by…
Question Number 162674 by Mathematification last updated on 31/Dec/21 Answered by mindispower last updated on 31/Dec/21 $${d}\mid\mathrm{2}\left(\mathrm{3}{n}+\mathrm{1}\right)−\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)\Rightarrow{d}\mid−\mathrm{1} \\ $$$${d}=\mathrm{1} \\ $$$${or}\:{use}\:{bizou}\:{identities} \\ $$$$−\mathrm{2}\left(\mathrm{3}{n}+\mathrm{1}\right)+\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{2}{n}+\mathrm{1}\right),\left(\mathrm{3}{n}+\mathrm{1}\right)\:{are}\:{prime}…
Question Number 31579 by Joel578 last updated on 10/Mar/18 $$\mathrm{Let}\:{a}\:\mathrm{and}\:{b}\:\mathrm{be}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{part}\:\mathrm{and}\:\mathrm{a}\:\mathrm{decimal} \\ $$$$\mathrm{fraction}\:\mathrm{of}\:\sqrt{\mathrm{7}},\:\mathrm{respectively}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{integer} \\ $$$$\mathrm{part}\:\mathrm{of}\:\frac{{a}}{{b}}\:\mathrm{is}? \\ $$ Answered by Tinkutara last updated on 10/Mar/18 $${a}=\mathrm{2} \\…
Question Number 96928 by bobhans last updated on 05/Jun/20 $$\mathrm{69}{x}\:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{31}\right)\: \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$ Commented by RAMANA last updated on 05/Jun/20 $${send}\:{the}\:{answer}\:{please} \\ $$ Answered…
Question Number 31125 by Joel578 last updated on 02/Mar/18 $$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}.\:\mathrm{Then}\:{x}^{\mathrm{2}} \:+\:\mathrm{1}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{factor}\:\mathrm{of}\:\left({x}^{\mathrm{4}} \:+\:\mathrm{3}\right)^{{n}} \:−\:\left[\left({x}^{\mathrm{2}} \:+\:\mathrm{3}\right)\left({x}^{\mathrm{2}} \:−\:\mathrm{1}\right)\right]^{{n}} \\ $$$$\mathrm{for}\:… \\ $$$$\left(\mathrm{A}\right)\:\mathrm{All}\:{n} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{Odd}\:{n} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{Even}\:{n}…
Question Number 162169 by Rasheed.Sindhi last updated on 27/Dec/21 $$ \\ $$$$\:\:\:\:\:\:\:\begin{array}{|c|}{\overset{\bullet} {\:\:\:\:\:\begin{array}{|c|}{\:\:\:\underset{{x}=?,{y}=?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {\overset{{x},{y}\in\mathbb{Z}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {{x}+{y}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{586}}}\:\:}\\\hline\end{array}_{} ^{} }\:\:\:\:}\\\hline\end{array} \\ $$$$ \\ $$ Answered by…
Question Number 96558 by bobhans last updated on 02/Jun/20 $$\mathrm{Let}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n}\:\mathrm{be}\:\mathrm{two}\:\mathrm{positive}\:\mathrm{integers}\: \\ $$$$\mathrm{satisfy}\:\frac{\mathrm{m}}{\mathrm{n}}\:=\:\frac{\mathrm{1}}{\mathrm{10}×\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{12}×\mathrm{14}}+\frac{\mathrm{1}}{\mathrm{14}×\mathrm{16}}+…+\frac{\mathrm{1}}{\mathrm{2012}×\mathrm{2014}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{m}+\mathrm{n}\: \\ $$ Answered by john santu last updated on…