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Question Number 75847 by behi83417@gmail.com last updated on 18/Dec/19 $$\begin{cases}{\frac{\boldsymbol{\mathrm{tgx}}−\boldsymbol{\mathrm{tgy}}}{\mathrm{1}−\boldsymbol{\mathrm{tgx}}.\boldsymbol{\mathrm{tgy}}}=\boldsymbol{\mathrm{tg}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}}\\{\:\:\frac{\boldsymbol{\mathrm{tgx}}+\boldsymbol{\mathrm{tgy}}}{\mathrm{1}+\boldsymbol{\mathrm{tgxtgy}}}=\boldsymbol{\mathrm{tg}}\frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}}}\end{cases} \\ $$ Commented by MJS last updated on 19/Dec/19 $$\mathrm{the}\:\mathrm{path}\:\mathrm{is} \\ $$$$\mathrm{let}\:{x}=\mathrm{2arctan}\:{p};\:{y}=\mathrm{2arctan}\:{q} \\ $$$$\Rightarrow \\…
Question Number 141383 by Willson last updated on 18/May/21 $$\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{find}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$ Answered by mr W last updated on 18/May/21 $$\mathrm{sin}^{\mathrm{2}}…
Question Number 10309 by Tawakalitu ayo mi last updated on 03/Feb/17 $$\mathrm{An}\:\mathrm{helium}\:\mathrm{atom}\:\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{6}.\mathrm{64}×\mathrm{10}^{−\mathrm{27}} \mathrm{kg} \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{charge}\:\mathrm{Q}\:\mathrm{is}\:+\mathrm{2}\:\mathrm{electron}.\:\mathrm{Compare}\:\mathrm{the} \\ $$$$\mathrm{magnitude}\:\mathrm{of}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{repultion}\:\mathrm{to}\:\mathrm{that} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{gravitational}\:\mathrm{attraction}\:\mathrm{between}\:\mathrm{them}. \\ $$ Terms of Service Privacy…
Question Number 75845 by behi83417@gmail.com last updated on 18/Dec/19 $$\begin{cases}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{yz}}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\\{\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{xz}}=\boldsymbol{\mathrm{y}}^{\mathrm{2}} }\\{\boldsymbol{\mathrm{z}}+\boldsymbol{\mathrm{xy}}=\boldsymbol{\mathrm{z}}^{\mathrm{2}} }\end{cases}\:\:\:\:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}. \\ $$ Commented by mr W last updated on 18/Dec/19 $${x}={y}={z}=\mathrm{0} \\…
Question Number 141378 by mnjuly1970 last updated on 18/May/21 $$\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\frac{\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)}{\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)}\:=\varphi\: \\ $$$$\:\:\:\:\:\:\:\varphi:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 141373 by SLVR last updated on 18/May/21 Commented by SLVR last updated on 18/May/21 $${kindly}\:{help}\:{me} \\ $$ Answered by mathmax by abdo last…
Question Number 75838 by Crabby89p13 last updated on 18/Dec/19 Commented by MJS last updated on 19/Dec/19 $$\mathrm{testing}\:\mathrm{all}\:\mathrm{derivates} \\ $$$$\Rightarrow \\ $$$$\frac{{d}}{{dx}}\left[\frac{{x}^{\mathrm{12}} }{\mathrm{6}\left(\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\right]=…=\frac{\mathrm{3}{x}^{\mathrm{13}}…
Question Number 10303 by amir last updated on 02/Feb/17 Commented by amir last updated on 02/Feb/17 $${triangle}\:{ABC}\:\:{is}\:{isoscale}.{BA}={BC} \\ $$$${find}\:\:\:\:\boldsymbol{{x}}. \\ $$ Terms of Service Privacy…
Question Number 141372 by iloveisrael last updated on 18/May/21 Answered by EDWIN88 last updated on 18/May/21 $$\left(\mathrm{1}\right)\:\mathrm{xy}\:+\:\frac{\mathrm{x}}{\mathrm{y}}\:=\:\mathrm{5} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{xy}−\frac{\mathrm{x}}{\mathrm{y}}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\begin{cases}{\mathrm{2xy}\:=\:\frac{\mathrm{26}}{\mathrm{5}}\rightarrow\mathrm{xy}=\frac{\mathrm{13}}{\mathrm{5}}}\\{\frac{\mathrm{2x}}{\mathrm{y}}\:=\:\frac{\mathrm{24}}{\mathrm{5}}\rightarrow\frac{\mathrm{x}}{\mathrm{y}}\:=\:\frac{\mathrm{12}}{\mathrm{5}}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{xy}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\:\frac{\mathrm{13}×\mathrm{12}}{\mathrm{25}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}}…