Question Number 210235 by peter frank last updated on 03/Aug/24 Commented by Pnk2024 last updated on 03/Aug/24 $$\bigtriangleup{ADB}\:\sim\bigtriangleup{EFB}\:\:\:\:….\:\left({A}−{A}\:{test}\right) \\ $$$$\Rightarrow\:\frac{{y}}{{x}}\:=\:\frac{{AB}}{{EB}}\:\:\:…….\:\left({C}.{S}.{S}.{T}\right) \\ $$$${again} \\ $$$$\bigtriangleup{BCA}\sim\bigtriangleup{EFA} \\…
Question Number 210234 by peter frank last updated on 03/Aug/24 Commented by Pnk2024 last updated on 03/Aug/24 $$\:{we}\:{know}\:{that} \\ $$$$\:\:{sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\Rightarrow\:\:{cos}^{\mathrm{2}} \theta=\mathrm{1}−{sin}^{\mathrm{2}}…
Question Number 210229 by mnjuly1970 last updated on 03/Aug/24 Answered by a.lgnaoui last updated on 03/Aug/24 $$\mathrm{8}=\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\mathrm{6}\boldsymbol{\mathrm{ab}} \\ $$$$ \\ $$$$\mathrm{16}=\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{4}\right)+\left(\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\mathrm{4}\right)+\mathrm{6}\boldsymbol{\mathrm{ab}}\:\:\:…
Question Number 210231 by a.lgnaoui last updated on 03/Aug/24 $$\mathrm{Resoudre}\:\boldsymbol{\mathrm{dans}}\:\mathbb{R} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{a}}\mathrm{cos}\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{b}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{c}}\:\:\:\:\:\left(\boldsymbol{\mathrm{x}}\neq\mathrm{0}\right)}\\{\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\boldsymbol{\mathrm{x}}}\right)\:\:\:\:\:\:\:\:\:=\boldsymbol{\mathrm{d}}\:\:\:\:\left(−\mathrm{1}\leqslant\boldsymbol{\mathrm{d}}\leqslant+\mathrm{1}\right)}\end{cases} \\ $$$$ \\ $$ Commented by mr W last updated on 04/Aug/24 $${they}\:{are}\:{two}\:{different}\:{equations}\:{for}…
Question Number 210206 by Spillover last updated on 02/Aug/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\frac{\mathrm{1}}{{e}}} ^{{e}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\mathrm{log}\:^{\mathrm{7}} {x}\right)} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 210208 by Spillover last updated on 02/Aug/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{{e}^{{e}^{{x}} } } \:{e}^{{e}^{{x}} } \:{e}^{{x}} {dx} \\ $$$$ \\ $$ Answered…
Question Number 210156 by Spillover last updated on 01/Aug/24 Answered by A5T last updated on 01/Aug/24 $$\left({i}\right)+\mathrm{2}×\left({ii}\right)\Rightarrow{x}=\frac{\mathrm{8}−\mathrm{7}{z}}{\mathrm{7}}=\frac{\mathrm{8}}{\mathrm{7}}−{z} \\ $$$$\left({iii}\right)+\left({i}\right)\Rightarrow\mathrm{7}{x}+\left({a}^{\mathrm{2}} −\mathrm{9}\right){z}={a}+\mathrm{4} \\ $$$$\Rightarrow{x}=\frac{−\left({a}^{\mathrm{2}} −\mathrm{9}\right){z}+{a}+\mathrm{4}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{\mathrm{4}−{a}}{\mathrm{7}}=\frac{\left(\mathrm{16}−{a}^{\mathrm{2}}…
Question Number 210142 by Abdullahrussell last updated on 01/Aug/24 Commented by Frix last updated on 02/Aug/24 $$\mathrm{I}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:“\mathrm{nice}''\:\mathrm{solution}: \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\:\:\:\:\:{y}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\:\:\:\:\:{z}=\mathrm{3} \\ $$$$\Rightarrow\:{x}+{y}+{z}=\mathrm{6} \\ $$$$\mathrm{But}\:\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{more}\:\mathrm{solutions}\:\mathrm{with} \\ $$$${x}+{y}+{z}\in\mathbb{R}…
Question Number 210157 by Spillover last updated on 01/Aug/24 Answered by A5T last updated on 01/Aug/24 $${Let}\:{distance}\:{of}\:{vertex},{V},\:{to}\:{centroid},{G},\:{be}\:{GV} \\ $$$$\Rightarrow\frac{{sin}\mathrm{30}°}{{GV}}=\frac{{sin}\mathrm{120}^{°} }{{x}}\Rightarrow{GV}=\frac{\frac{{x}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${H}=\sqrt{{x}^{\mathrm{2}} −{GV}^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}}…
Question Number 210155 by Spillover last updated on 01/Aug/24 Answered by A5T last updated on 01/Aug/24 Commented by A5T last updated on 01/Aug/24 $${C}\:{is}\:{the}\:{origin}\:{and}\:{redundant} \\…