Question Number 132597 by Study last updated on 15/Feb/21 Commented by mr W last updated on 15/Feb/21 $$\frac{\mathrm{156}}{{x}+\mathrm{7}}=\frac{\mathrm{156}}{{x}}−\mathrm{14} \\ $$$$\mathrm{156}{x}=\mathrm{156}\left({x}+\mathrm{7}\right)−\mathrm{14}{x}\left({x}+\mathrm{7}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{78}=\mathrm{0} \\ $$$$\left({x}−\mathrm{6}\right)\left({x}+\mathrm{13}\right)=\mathrm{0}…
Question Number 132594 by Study last updated on 15/Feb/21 Commented by mr W last updated on 15/Feb/21 $${the}\:{sum}\:{of}\:{two}\:{complex}\:{numbers} \\ $$$${can}\:{be}\:{a}\:{real}\:{number},\:{e}.{g}.\: \\ $$$$\left(\mathrm{1}+{i}\right)+\left(\mathrm{1}−{i}\right)=\mathrm{2} \\ $$$${the}\:{product}\:{of}\:{two}\:{complex}\:{numbers} \\…
Question Number 132575 by nadovic last updated on 15/Feb/21 $$\mathrm{Hello}\:\mathrm{Tinku}\:\mathrm{Tara},\:\mathrm{I}\:\mathrm{have}\:\mathrm{a}\:\mathrm{new}\:\mathrm{phone} \\ $$$$\mathrm{and}\:\mathrm{I}\:\mathrm{am}\:\mathrm{unable}\:\mathrm{to}\:\mathrm{remember}\:\mathrm{the}\: \\ $$$$\mathrm{password}\:\mathrm{for}\:\mathrm{my}\:\mathrm{previous}\:\mathrm{account}.\:{Is} \\ $$$${there}\:{anyway}\:{to}\:{help}\:{me}\:{retrieve}\:{it}? \\ $$ Commented by mr W last updated on…
Question Number 67031 by hmamarques1994@gmai.com last updated on 21/Aug/19 $$\: \\ $$$$\:\underset{\boldsymbol{\mathrm{x}}\rightarrow−\mathrm{1}} {\boldsymbol{\mathrm{lim}}}\sqrt[{\mathrm{3}}]{\frac{\sqrt[{\mathrm{7}}]{\boldsymbol{\mathrm{x}}^{\mathrm{5}} }+\mathrm{1}}{\mathrm{1}+\sqrt[{\mathrm{9}\:}]{\boldsymbol{\mathrm{x}}^{\mathrm{7}} }}}=? \\ $$$$\: \\ $$ Commented by Tony Lin last updated…
Question Number 132564 by abony1303 last updated on 15/Feb/21 Answered by abony1303 last updated on 15/Feb/21 $$\mathrm{pls}\:\mathrm{help}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{region} \\ $$ Answered by MJS_new last updated on…
Question Number 1486 by 112358 last updated on 13/Aug/15 $$\mathrm{While}\:\mathrm{this}\:\mathrm{may}\:\mathrm{be}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{off}\: \\ $$$$\mathrm{the}\:\mathrm{usual}\:\mathrm{Q}\:\mathrm{n}\:\mathrm{A}\:\mathrm{posts}\:\mathrm{here}, \\ $$$$\mathrm{what}\:\mathrm{advice}\:\mathrm{can}\:\mathrm{be}\:\mathrm{given}\:\mathrm{to}\: \\ $$$$\mathrm{someone}\:\mathrm{wishing}\:\mathrm{to}\:\mathrm{successfully} \\ $$$$\mathrm{carry}\:\mathrm{out}\:\mathrm{undergraduate}\:\mathrm{studies} \\ $$$$\mathrm{in}\:\mathrm{mathematics}?\:\mathrm{I}'\mathrm{m}\:\mathrm{worried}\: \\ $$$$\mathrm{about}\:\mathrm{my}\:\mathrm{problem}\:\mathrm{solving}\:\mathrm{skills} \\ $$$$\mathrm{and}\:\mathrm{my}\:\mathrm{ability}\:\mathrm{to}\:\mathrm{consider}\:\mathrm{all} \\…
Question Number 132553 by rs4089 last updated on 15/Feb/21 Commented by MJS_new last updated on 15/Feb/21 $$\mathrm{use}\:\mathrm{software}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{it} \\ $$ Answered by guyyy last updated on…
Question Number 132554 by mohammad17 last updated on 15/Feb/21 Commented by MJS_new last updated on 15/Feb/21 $$\mathrm{we}\:\mathrm{had}\:\mathrm{this}\:\mathrm{many}\:\mathrm{times}\:\mathrm{before}.\:\mathrm{simply} \\ $$$$\mathrm{substitute}\:{t}=\sqrt{\mathrm{tan}\:\theta}\:\mathrm{and}\:\mathrm{there}\:\mathrm{you}\:\mathrm{go} \\ $$ Commented by liberty last…
Question Number 132528 by greg_ed last updated on 14/Feb/21 $$\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{each}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{brings}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{forum}}\:! \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 1453 by 112358 last updated on 06/Aug/15 $${Show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{rx}} =\frac{{e}^{{x}} \left({e}^{{nx}} −\mathrm{1}\right)}{{e}^{{x}} −\mathrm{1}}\:\:\:\:\:\:\:\left(\ast\right) \\ $$$${if}\:\:\:{e}^{{rx}} ={cos}\left({irx}\right)−{isin}\left({irx}\right)\:{and}\:{x}\neq\mathrm{0}\:. \\ $$$$\left[{Do}\:{not}\:{treat}\:\left(\ast\right)\:{as}\:{a}\:{GP}\:{to}\right. \\ $$$$\left.{directly}\:{obtain}\:{the}\:{result}.\right]…