Question Number 180207 by Acem last updated on 09/Nov/22 $$\:\frac{\mathrm{2}}{\mathrm{15}}\:,\:\:\frac{\mathrm{11}}{\mathrm{40}}\:\:,\:\:\frac{\mathrm{26}}{\mathrm{75}}\:\:,\:\:\frac{\mathrm{47}}{\mathrm{120}}\:\:,\:… \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated on 09/Nov/22 $${Numerators}: \\ $$$$\begin{bmatrix}{\mathrm{2}}&{\:}&{\mathrm{11}}&{\:}&{\mathrm{26}}&{\:}&{\mathrm{47}}&{\:}&{\mathrm{74}}\\{\:}&{\mathrm{9}}&{\:}&{\mathrm{15}}&{\:}&{\mathrm{21}}&{\:}&{\mathrm{27}}&{\:}\\{\:}&{\:}&{\:\mathrm{6}}&{\:}&{\mathrm{6}}&{\:}&{\mathrm{6}}&{\:}&{\:}\end{bmatrix}\:\: \\…
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Question Number 49123 by Pk1167156@gmail.com last updated on 03/Dec/18 $$\mathrm{If} \\ $$$$\mathrm{a}^{\mathrm{3}} −\mathrm{a}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{valueof} \\ $$$$\mathrm{a}^{\mathrm{4}} +\mathrm{a}^{\mathrm{3}} −\mathrm{a}^{\mathrm{2}} −\mathrm{2a}+\mathrm{1} \\ $$ Commented by behi83417@gmail.com…
Question Number 180188 by Shrinava last updated on 08/Nov/22 Answered by Rasheed.Sindhi last updated on 09/Nov/22 $$\mathcal{F}{ormulas}: \\ $$$$\begin{array}{|c|}{\:\:\begin{bmatrix}{\:\:\:\:\:{l}}&{{m}}\\{−{m}}&{{l}}\end{bmatrix}\begin{bmatrix}{\:\:\:\:{p}}&{{q}}\\{−{q}}&{{p}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\:\:\:{lp}−{mq}}&{{lq}+{mp}}\\{−\left({lq}+{mp}\right)}&{{lp}−{mq}}\end{bmatrix}_{\:} ^{\:} }\\\hline\end{array}\: \\ $$$$\begin{array}{|c|}{\left(\begin{bmatrix}{\:\:\:\:\:{l}}&{{m}}\\{−{m}}&{{l}}\end{bmatrix}\right)^{\mathrm{2}} \:=\begin{bmatrix}{{l}^{\mathrm{2}} −{m}^{\mathrm{2}}…
Question Number 114653 by bemath last updated on 20/Sep/20 $${solve}\:\mathrm{6}{x}^{\mathrm{4}} −\mathrm{25}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{6}=\mathrm{0} \\ $$ Answered by bobhans last updated on 20/Sep/20 $$\Leftrightarrow\mathrm{6}\left({x}^{\mathrm{4}} +\mathrm{1}\right)−\mathrm{25}\left({x}^{\mathrm{3}} +{x}\right)+\mathrm{12}{x}^{\mathrm{2}}…
Question Number 114625 by bobhans last updated on 20/Sep/20 Answered by john santu last updated on 20/Sep/20 $$\Leftrightarrow{k}^{\mathrm{4}} +\mathrm{2}{k}^{\mathrm{2}} +\mathrm{9}={k}^{\mathrm{4}} +\mathrm{6}{k}^{\mathrm{2}} +\mathrm{9}−\mathrm{4}{k}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{3}\right)\left({k}^{\mathrm{2}}…
Question Number 180149 by Shrinava last updated on 07/Nov/22 $$\mathrm{If}\:\:\omega−\mathrm{Brocard}'\mathrm{s}\:\mathrm{angle}\:\mathrm{in}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{then}: \\ $$$$\underset{\boldsymbol{\mathrm{cyc}}} {\prod}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\omega}\:−\:\mathrm{2}\:\mathrm{cos}\:\mathrm{A}\right)\:\geqslant\:\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 114592 by Aziztisffola last updated on 19/Sep/20 $$\mathrm{Solve}\::\:{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$ Answered by MJS_new last updated on 19/Sep/20 $${x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}=\mathrm{3}{k}\:\mathrm{with}\:{k}\in\mathbb{Z} \\ $$$$\Rightarrow\:{x}=−\mathrm{1}\pm\sqrt{\mathrm{3}{k}+\mathrm{2}} \\…