Question Number 132102 by mr W last updated on 14/Feb/21 Commented by mr W last updated on 16/Feb/21 $${a}\:{ball}\:{is}\:{thrown}\:{from}\:{point}\:{A}\:{with} \\ $$$${speed}\:\boldsymbol{{u}}\:{and}\:{strikes}\:{at}\:{a}\:{point}\:{B} \\ $$$${on}\:{the}\:{semispherical}\:{surface}\:{with} \\ $$$${radius}\:{R}\:{and}\:{returns}\:{back}\:{to}\:{point}\:{A}.…
Question Number 66472 by hmamarques1994@gmail.com last updated on 15/Aug/19 $$\begin{cases}{\sqrt[{\sqrt{\mathrm{6}}}]{\boldsymbol{\mathrm{x}}}+\sqrt[{\sqrt{\mathrm{5}}}]{\boldsymbol{\mathrm{y}}}=\mathrm{11}}\\{\frac{\sqrt[{\sqrt{\mathrm{5}}}]{\boldsymbol{\mathrm{y}}}}{\:\sqrt[{\sqrt{\mathrm{6}}}]{\boldsymbol{\mathrm{x}}}}=\mathrm{1}\frac{\mathrm{1}}{\mathrm{5}}}\end{cases} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Qual}}\:\:\acute {\boldsymbol{\mathrm{e}}}\:\:\boldsymbol{\mathrm{o}}\:\:\boldsymbol{\mathrm{par}}\:\:\boldsymbol{\mathrm{ordenado}}\:\:\boldsymbol{\mathrm{na}}\:\:\boldsymbol{\mathrm{forma}}\:\:\boldsymbol{\mathrm{a}}^{\sqrt{\boldsymbol{\mathrm{p}}}} \:\:\boldsymbol{\mathrm{e}}\:\:\boldsymbol{\mathrm{b}}^{\sqrt{\boldsymbol{\mathrm{q}}}} \\ $$$$\:\boldsymbol{\mathrm{que}}\:\:\boldsymbol{\mathrm{satisfaz}}\:\:\boldsymbol{\mathrm{o}}\:\:\boldsymbol{\mathrm{sistema}}\:\:\boldsymbol{\mathrm{como}}\:\:\boldsymbol{\mathrm{possivel}}\:\:\boldsymbol{\mathrm{e}}\:\:\boldsymbol{\mathrm{determinado}}? \\ $$ Answered by MJS last updated…
Question Number 651 by 123456 last updated on 19/Feb/15 $$\mathrm{arg}\left(\mathrm{z}−\mathrm{a}\right)−\mathrm{arg}\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)−\mathrm{arg}\left(\mathrm{z}−\bar {\mathrm{z}}_{\mathrm{1}} \right)={k}\pi \\ $$$${a}\in\mathbb{R} \\ $$$${z}_{\mathrm{1}} \in\mathbb{C},\Im\left(\bar {{z}}_{\mathrm{1}} \right)\neq\mathrm{0} \\ $$$${z}\in\mathbb{C} \\ $$$${k}\in\mathbb{Z} \\…
Question Number 132102 by mr W last updated on 14/Feb/21 Commented bymr W last updated on 16/Feb/21 $${a}\:{ball}\:{is}\:{thrown}\:{from}\:{point}\:{A}\:{with} \\ $$ $${speed}\:\boldsymbol{{u}}\:{and}\:{strikes}\:{at}\:{a}\:{point}\:{B} \\ $$ $${on}\:{the}\:{semispherical}\:{surface}\:{with} \\…
Question Number 66562 by Rio Michael last updated on 17/Aug/19 $${given}\:{that}\:\:\mid{z}\:−\:\mathrm{i}\mid\:=\:\mid{z}\:−\:\mathrm{4}\:+\mathrm{3}\:\mathrm{i}\mid \\ $$$${sketch}\:{the}\:{locus}\:{of}\:\:{z} \\ $$$${find}\:{the}\:{catersian}\:{equation}\:{of}\:{this}\:{locus}. \\ $$ Commented by mathmax by abdo last updated on…
Question Number 132098 by rs4089 last updated on 11/Feb/21 Answered by Ar Brandon last updated on 11/Feb/21 $$\mathrm{S}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} +\mathrm{kn}+\mathrm{k}^{\mathrm{2}} } \\ $$$$\underset{\mathrm{n}\rightarrow\infty}…
Question Number 66561 by Rio Michael last updated on 17/Aug/19 $${evaluate}\: \\ $$$$\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \mid\:{x}+\:\mathrm{2}\mid\:{dx}. \\ $$ Commented by kaivan.ahmadi last updated on 17/Aug/19 $$\mathrm{0}<{x}<\mathrm{2}\Rightarrow{x}+\mathrm{2}>\mathrm{0}\Rightarrow\mid{x}+\mathrm{2}\mid={x}+\mathrm{2}…
Question Number 1024 by brahim last updated on 19/May/15 $$\mathrm{4}{x}+\mathrm{5}=\mathrm{6}−\mathrm{4}\left(\mathrm{1}−{x}\right) \\ $$ Answered by prakash jain last updated on 19/May/15 $$\mathrm{4}{x}+\mathrm{5}=\mathrm{6}−\mathrm{4}+\mathrm{4}{x} \\ $$$$\mathrm{Equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}. \\ $$…
Question Number 1023 by 123456 last updated on 20/May/15 $$\phi_{{n}} :\mathbb{R}\rightarrow\mathbb{R} \\ $$$${n}\in\mathbb{N}^{\ast} \\ $$$$\phi_{{n}} ={t}^{{n}} \frac{{d}^{{n}} \phi}{{dt}^{{n}} } \\ $$$$\phi_{\mathrm{1}} \left({t}\right)=?,\phi_{\mathrm{1}} \left(\mathrm{1}\right)=+\mathrm{1} \\ $$$$\phi_{\mathrm{2}}…
Question Number 132089 by rs4089 last updated on 11/Feb/21 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$ Answered by Lordose last updated on 11/Feb/21 $$\Omega\:=\:\int_{\mathrm{0}}…