Question Number 41389 by mondodotto@gmail.com last updated on 06/Aug/18 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\mathrm{B}}+\mathrm{3}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{B}}+\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{A}}=\mathrm{0} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 06/Aug/18…
Question Number 172455 by pablo1234523 last updated on 27/Jun/22 $$\mathrm{using}\:\mathrm{maclaurin}'\mathrm{s}\:\mathrm{series}\:\mathrm{exapand} \\ $$$$\frac{{x}}{\mathrm{2}}\left(\frac{{e}^{{x}} +\mathrm{1}}{{e}^{{x}} −\mathrm{1}}\right)\:\mathrm{upto}\:{x}^{\mathrm{4}} \:\mathrm{term} \\ $$ Answered by mr W last updated on 29/Jun/22…
Question Number 172443 by SANOGO last updated on 27/Jun/22 Answered by Mathspace last updated on 27/Jun/22 $$\left.\mathrm{2}\right)\:{la}\:{fonctionx}\rightarrow\:\Upsilon\left({x},{t}\right)\:{est}\:{derivable}\:{surR} \\ $$$${et}\:\frac{\partial}{\partial{x}}\Psi\left({x},{t}\right)=\left({e}^{−{at}} −{e}^{−{bt}} \right){cos}\left({xt}\right) \\ $$$${t}\rightarrow\frac{\left({e}^{−{at}} −{e}^{−{bt}} \right){cos}\left({xt}\right)}{{t}}\:{est}\:{integrable}…
Question Number 172439 by Physicien last updated on 27/Jun/22 Commented by mr W last updated on 27/Jun/22 $${this}\:{app}\:{Equation}\:{Editor}\:{is}\:{made}\:{for} \\ $$$${writing}\:{equations}.\:{Why}\:{don}'{t}\:{you} \\ $$$${use}\:{it}\:{for}\:{writing}\:{equations}\:/\:{formulas} \\ $$$${instead}\:{of}\:{posting}\:{pictures}?\:{is}\:{it} \\…
Question Number 172420 by Physicien last updated on 26/Jun/22 Answered by thfchristopher last updated on 26/Jun/22 $$\int\frac{\mathrm{1}}{\mathrm{7}+\mathrm{tan}\:{x}}{dx} \\ $$$$\mathrm{Let}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$${dt}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \:{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dc}…
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Question Number 106847 by mathocean1 last updated on 07/Aug/20 $${Given}\:{f}\left({x}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{{sinx}}+\frac{\mathrm{1}}{{cosx}} \\ $$$${show}\:{that}\:{f}\:'\left({x}\right)={cosx}\frac{\left({tan}^{\mathrm{3}} {x}−\mathrm{3}\sqrt{\mathrm{3}}\right)}{{sin}^{\mathrm{2}} {x}} \\ $$ Answered by bemath last updated on 07/Aug/20 $$\:\:\:\:\:^{@\mathrm{bemath}@} \\…
Question Number 41292 by behi83417@gmail.com last updated on 04/Aug/18 Answered by MJS last updated on 04/Aug/18 $${m}=\frac{{n}+\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{−\mathrm{3}{n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{root}\:\mathrm{is}\:\mathrm{only}\:\mathrm{defined}\:\mathrm{for}\:{n}=\mathrm{0},\:\mathrm{1},\:\mathrm{2}\:\mathrm{and}\:\mathrm{for} \\ $$$$\mathrm{these}\:\mathrm{values}\:{m}\notin\mathbb{N} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\…
Question Number 41288 by mondodotto@gmail.com last updated on 04/Aug/18 $$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{knowledge}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{series}}\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{compound}}\:\boldsymbol{\mathrm{interest}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{by}} \\ $$$$\boldsymbol{\mathrm{An}}=\boldsymbol{\mathrm{P}}\left(\mathrm{1}+\frac{\boldsymbol{\mathrm{RT}}}{\mathrm{100}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18…
Question Number 172326 by nurtani last updated on 25/Jun/22 $${If}\: \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}\right)+\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}\right)\:=\:\mathrm{70} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:=\:\:? \\ $$ Commented by infinityaction…