Question Number 138702 by peter frank last updated on 16/Apr/21 Answered by mr W last updated on 17/Apr/21 $${y}=\mathrm{sinh}\:{x}+{k}\:\mathrm{cosh}\:{x} \\ $$$$\:\:=\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\:\left(\frac{\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}}\mathrm{sinh}\:{x}+\frac{{k}}{\:\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{cosh}\:{x}\right) \\…
Question Number 7612 by Tawakalitu. last updated on 06/Sep/16 $${If}\:\:{ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{0},\: \\ $$$${and},\:\:{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} {y}\:+\:{c}^{\mathrm{2}} {z}\:=\:\mathrm{0} \\ $$$${Find}\:{the}\:{ratio}\:\:{x}:{y}:{z} \\ $$ Commented by Rasheed Soomro last updated…
Question Number 7611 by Tawakalitu. last updated on 06/Sep/16 $${Find}\:{the}\:{square}\:{root}\:{of}\: \\ $$$$\mathrm{121}{x}^{\mathrm{6}} \:+\:\mathrm{44}{x}^{\mathrm{5}} \:−\:\mathrm{18}{x}^{\mathrm{4}} \:+\:\mathrm{18}{x}^{\mathrm{3}} \:+\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$ Answered by Rasheed Soomro last updated…
Question Number 7610 by Tawakalitu. last updated on 06/Sep/16 $${obtain}\:{the}\:{value}\:{of}\: \\ $$$$\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} −{a}^{\mathrm{1}/\mathrm{2}} \right)\: \\…
Question Number 73131 by behi83417@gmail.com last updated on 06/Nov/19 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\:\:\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\:. \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}^{\mathrm{2}} \\ $$ Commented by mr W last updated on 06/Nov/19 $${x}\geqslant\mathrm{0}…
Question Number 7598 by Rohit last updated on 05/Sep/16 $${solve}\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\mid<\mathrm{3} \\ $$ Answered by Rasheed Soomro last updated on 06/Sep/16 $$\mid\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\mid<\mathrm{3}…
Question Number 73113 by TawaTawa last updated on 06/Nov/19 Commented by mathmax by abdo last updated on 06/Nov/19 $$\left.\mathrm{1}\right)\:{we}\:{have}\:{arg}\left({z}\right)={arg}\left(\mathrm{7}−\mathrm{3}\sqrt{\mathrm{3}}{i}\right)+{arg}\left(−\mathrm{1}−{i}\right)\left[\mathrm{2}\pi\right] \\ $$$$\mid\mathrm{7}−\mathrm{3}\sqrt{\mathrm{3}}\mid\:=\sqrt{\mathrm{49}+\mathrm{27}}=\sqrt{\mathrm{76}}\:\Rightarrow\mathrm{7}−\mathrm{3}\sqrt{\mathrm{3}}=\sqrt{\mathrm{76}}{e}^{{iarctan}\left(\frac{−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{7}}\right)} \:\Rightarrow \\ $$$${arg}\left(\mathrm{7}−\mathrm{3}\sqrt{\mathrm{3}}\right)\:=−{arctan}\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{7}}\right) \\…
Question Number 7577 by Tawakalitu. last updated on 04/Sep/16 $${p}\:=\:\frac{{a}\left[\mathrm{1}\:−\:\left(\mathrm{1}\:+\:{r}\right)^{−{n}} \right]}{{r}} \\ $$$$ \\ $$$${Make}\:\:\:{r}\:\:\:{the}\:{subject}\:{of}\:{the}\:{fomular}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 138641 by soudo last updated on 15/Apr/21 Answered by MJS_new last updated on 16/Apr/21 $$\left.\mathrm{f}\left.\mathrm{rom}\:\mathrm{1}\right)\:\Rightarrow\:\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{x}^{\mathrm{2}} \pm\mathrm{2}{xy}+{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=…
Question Number 7532 by Yozzia last updated on 02/Sep/16 $${Find}\:{x}_{{n}} \:\left({n}\in\mathbb{Z}\right)\:{satisfying}\:{x}_{\mathrm{0}} =\mathrm{0},\:{x}_{\mathrm{1}} =\mathrm{1}\:{and} \\ $$$${x}_{{n}+\mathrm{1}} ={x}_{{n}} \sqrt{{x}_{{n}−\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}}+{x}_{{n}−\mathrm{1}} \sqrt{{x}_{{n}} ^{\mathrm{2}} +\mathrm{1}}\:{for}\:{n}\geqslant\mathrm{1}. \\ $$ Commented…