Question Number 49604 by Tawa1 last updated on 08/Dec/18 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\mathrm{a}\:+\:\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\leqslant\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \\ $$ Answered by afachri last updated on 08/Dec/18 $$\mathrm{let}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}\:−\:{b}\right)^{\mathrm{2}} \:\:\geqslant\:\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}}…
Question Number 115135 by Algoritm last updated on 23/Sep/20 Commented by Dwaipayan Shikari last updated on 23/Sep/20 $$\frac{\mathrm{1}}{\mathrm{51}.\mathrm{52}…\mathrm{100}}=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}..}{\mathrm{100}!}=\frac{\mathrm{50}!}{\mathrm{100}!} \\ $$$$\mathrm{1}.\mathrm{3}.\mathrm{5}..\mathrm{9}…=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}…}{\mathrm{2}^{\mathrm{100}} .\mathrm{50}!} \\ $$$$\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{50}} .\mathrm{50}!}.\frac{\mathrm{50}!}{\mathrm{100}!}=\mathrm{2}^{\mathrm{x}} \\…
Question Number 115136 by Algoritm last updated on 23/Sep/20 Answered by Olaf last updated on 23/Sep/20 $$\mathrm{sin2}\theta\:=\:\mathrm{2sin}\theta\mathrm{cos}\theta \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{sin2}\theta}{\mathrm{sin}\theta} \\ $$$$\theta\:=\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}},\:\mathrm{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{sin}\left(\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}}…
Question Number 180653 by Shrinava last updated on 14/Nov/22 Answered by aleks041103 last updated on 17/Nov/22 $${A}=\begin{pmatrix}{{a}}&{{b}}\\{−{b}}&{{a}}\end{pmatrix}\:={a}\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\:+{b}\begin{pmatrix}{\mathrm{0}}&{\mathrm{1}}\\{−\mathrm{1}}&{\mathrm{0}}\end{pmatrix}\:={aE}+{bS} \\ $$$${AB}=\left({aE}+{bS}\right)\left({cE}+{dS}\right)= \\ $$$$={acE}+{bdS}^{\mathrm{2}} +\left({ad}+{bc}\right){S} \\ $$$${S}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{0}}&{\mathrm{1}}\\{−\mathrm{1}}&{\mathrm{0}}\end{pmatrix}\begin{pmatrix}{\mathrm{0}}&{\mathrm{1}}\\{−\mathrm{1}}&{\mathrm{0}}\end{pmatrix}\:=\:\begin{pmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{1}}\end{pmatrix}\:=−{E}…
Question Number 180652 by Shrinava last updated on 14/Nov/22 Answered by aleks041103 last updated on 17/Nov/22 $${let}\:{detX}={x}\Rightarrow{a}=\mathrm{1} \\ $$$${det}\left({A}^{\mathrm{2}} \right)=\left({detA}\right)^{\mathrm{2}} =\mathrm{1}={det}\left({BC}\right)={detBdetC} \\ $$$$\Rightarrow{bc}=\mathrm{1}\Rightarrow{b}=\frac{\mathrm{1}}{{c}} \\ $$$${B}^{\mathrm{2}}…
Question Number 180641 by mr W last updated on 14/Nov/22 $${if}\:{a}+{b}+{c}+{d}+{e}=\mathrm{8}\:{and} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} =\mathrm{16},\:{what}\:{is}\:{the} \\ $$$${maximal}\:{value}\:{of}\:{a}\:? \\ $$ Answered by Emrice…
Question Number 49570 by Tawa1 last updated on 07/Dec/18 $$\mathrm{Eliminate}\:\:\boldsymbol{\mathrm{t}}\:\:\mathrm{from}\:\mathrm{this}\:\mathrm{equation}:\:\:\left(\mathrm{1}\right)\:\:\:\mathrm{x}\:=\:\mathrm{1}\:+\:\mathrm{t},\:\:\:\mathrm{y}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{t}} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{x}\:=\:\mathrm{3}\:+\:\mathrm{t}^{\mathrm{3}} \:,\:\:\:\:\:\mathrm{y}\:=\:\mathrm{2}\:+\:\frac{\mathrm{1}}{\mathrm{t}} \\ $$ Answered by afachri last updated on 07/Dec/18 $$\left.\mathrm{1}\right)\:\:{t}\:=\:{x}\:−\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\mathrm{so},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}\:=\:\mathrm{1}\:+\:\frac{\:\mathrm{1}\:}{\:{x}\:−\:\mathrm{1}\:}…
Question Number 49555 by ggny last updated on 07/Dec/18 $$\mathrm{Find}\:\mathrm{4}\:\: \\ $$$$\mathrm{plz}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sir} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 180604 by mathlove last updated on 14/Nov/22 Answered by Frix last updated on 14/Nov/22 $${x}=\sqrt{\mathrm{2}}\:^{\sqrt{\mathrm{2}}\:^{\sqrt{\mathrm{2}}\:^{…} } } =\sqrt{\mathrm{2}}\:^{{x}} \\ $$$$\mathrm{ln}\:{x}\:={x}\mathrm{ln}\:\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{ln}\:{x}}{{x}}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\…
Question Number 180603 by mr W last updated on 14/Nov/22 $${find}\:{the}\:{real}\:{solution}\:{of}\:{following} \\ $$$${equation}\:{system}: \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{xy}}+\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{p}} \\ $$$$\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{yz}}+\boldsymbol{{z}}^{\mathrm{2}} =\boldsymbol{{q}} \\ $$$$\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{zx}}+\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{r}}…