Question Number 69966 by necxxx last updated on 29/Sep/19 Commented by necxxx last updated on 29/Sep/19 $${please}\:{help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 4430 by alib last updated on 24/Jan/16 $$\left\{\left({x}−{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} \right. \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{15}{a}^{\mathrm{2}} \\ $$$$ \\ $$$${Solve}\:{the}\:{system}\:{of}\:{equation} \\ $$$$ \\ $$…
Question Number 4429 by alib last updated on 24/Jan/16 $$\left({sin}\:{x}\:+\sqrt{\mathrm{3}}\:{cos}\:{x}\right)\:{sin}\:\mathrm{3}{x}\:=\:\mathrm{2} \\ $$ Answered by Yozzii last updated on 24/Jan/16 $${sinx}+\sqrt{\mathrm{3}}{cosx}=\sqrt{\mathrm{1}+\mathrm{3}}{sin}\left({x}+\pi/\mathrm{3}\right) \\ $$$$=\mathrm{2}{sin}\left({x}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\left({sinx}+\sqrt{\mathrm{3}}{cosx}\right){sin}\mathrm{3}{x}=\mathrm{2}………\left(\Upsilon\right) \\…
Question Number 135497 by EDWIN88 last updated on 13/Mar/21 $$\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\:=\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)+\:\sqrt{\mathrm{7}}\:\mathrm{cos}\:\mathrm{x} \\ $$ Answered by john_santu last updated on 13/Mar/21 $$\Leftrightarrow\:\mathrm{sin}\:^{\mathrm{2}} \left({x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{sin}\:^{\mathrm{2}} \left({x}−\frac{\pi}{\mathrm{4}}\right)=\sqrt{\mathrm{7}}\:\mathrm{cos}\:{x} \\…
Question Number 135498 by aurpeyz last updated on 13/Mar/21 $$ \\ $$$$\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}{dx} \\ $$ Answered by john_santu last updated on 13/Mar/21 $$\mathscr{F}\:=\:\int\:\frac{{dx}}{{x}^{\mathrm{3}} \:\sqrt{\mathrm{25}{x}^{−\mathrm{2}}…
Question Number 135493 by EDWIN88 last updated on 13/Mar/21 $$\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{2}}\:=\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}} \\ $$ Answered by john_santu last updated on 13/Mar/21 $$\left(\mathrm{1}\right)\begin{cases}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2}\geqslant\mathrm{0}}\\{{x}^{\mathrm{2}} −{x}−\mathrm{2}\geqslant\mathrm{0}}\\{{x}^{\mathrm{2}}…
Question Number 4423 by Rasheed Soomro last updated on 24/Jan/16 $$\mathrm{Determine}\:\mathrm{f}\left(\mathrm{t}\right)\:\mathrm{such}\:\mathrm{that}\:\forall\:\mathrm{t}\in\mathbb{Z} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{8x}+\mathrm{f}\left(\mathrm{t}\right)\:\mathrm{has}\:\mathrm{integer}-\mathrm{solution}. \\ $$ Commented by Yozzii last updated on 24/Jan/16 $${Try}\:{f}\left({t}\right)={c};\:{c}\:{is}\:{a}\:{real}\:{constant}. \\…
Question Number 135495 by greg_ed last updated on 13/Mar/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{guyz}}\:! \\ $$$$\boldsymbol{\mathrm{let}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{try}}\:\boldsymbol{\mathrm{this}}\::\:\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{x}}}\boldsymbol{{dx}}. \\ $$ Answered by mathmax by abdo last updated…
Question Number 4420 by ankitbhawarkar333gmail.com last updated on 24/Jan/16 Answered by Yozzii last updated on 24/Jan/16 $${f}\left({x}\right)={sin}\mathrm{3}{x}−\mathrm{3}{sinx} \\ $$$${f}^{'} \left({x}\right)=\mathrm{3}{cos}\mathrm{3}{x}−\mathrm{3}{cosx} \\ $$$$\therefore\:{f}^{'} \left(\pi/\mathrm{2}\right)=\mathrm{3}{cos}\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{3}{cos}\frac{\pi}{\mathrm{2}}=\mathrm{0}. \\ $$$${Since}\:{f}^{'}…
Question Number 69954 by 20190927 last updated on 29/Sep/19 $$\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\:\mathrm{find}\:\mathrm{A}^{\mathrm{n}} \\ $$ Commented by mathmax by abdo last updated on 29/Sep/19 $${A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:={I}\:+{J} \\ $$$${we}\:{have}\:{J}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:.\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}…