Question Number 208681 by Noorzai last updated on 20/Jun/24 Answered by Rasheed.Sindhi last updated on 21/Jun/24 $$\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\:=\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}\:={a}+{b}\sqrt{\mathrm{3}}\:\left(>\mathrm{0}\right)\left({let}\right) \\ $$$$\:{Where}\:{a},{b}\in\mathbb{Z} \\ $$$$\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:={a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{\mathrm{3}}\: \\ $$$${a}^{\mathrm{2}}…
Question Number 208421 by lepuissantcedricjunior last updated on 15/Jun/24 Commented by Frix last updated on 15/Jun/24 $$\Psi=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{tan}\:\mathrm{2}{x}}}\:\overset{{t}=\sqrt{\mathrm{tan}\:\mathrm{2}{x}}} {=}\:\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{1}}=\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{2}} \\ $$…
Question Number 208420 by lepuissantcedricjunior last updated on 15/Jun/24 $$\boldsymbol{{resoudre}}\:\boldsymbol{{dans}}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\begin{cases}{\boldsymbol{{x}}+\boldsymbol{{y}}=\mathrm{3}}\\{\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{5}}\end{cases}\boldsymbol{{x}}+\boldsymbol{{z}}=\mathrm{4} \\ $$ Commented by A5T last updated on 15/Jun/24 $${You}\:{should}\:{learn}\:{to}\:{signify}\:{that}\:{you}\:{changed} \\ $$$${a}\:{question}. \\…
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Question Number 208178 by emilagazade last updated on 07/Jun/24 Commented by emilagazade last updated on 07/Jun/24 $${please}\:{help} \\ $$ Commented by A5T last updated on…
Question Number 207834 by efronzo1 last updated on 28/May/24 $$\:\:\:\:\underbrace{\:} \\ $$$$ \\ $$ Answered by Berbere last updated on 28/May/24 $${a}^{\mathrm{505}} ={x};{y}={b}^{\mathrm{505}} \\ $$$$\Leftrightarrow\begin{cases}{{x}+{y}.}\\{{max}\left\{{x}^{\mathrm{4}}…
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Question Number 207457 by efronzo1 last updated on 16/May/24 $$\:\:\:\cancel{\boldsymbol{{B}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 207381 by efronzo1 last updated on 13/May/24 Answered by mr W last updated on 13/May/24 $$\frac{{BD}\centerdot{CE}}{{AC}\centerdot{DE}} \\ $$$$=\frac{\left({BE}+{DE}\right)\centerdot{CE}}{\left({AE}+{CE}\right)\centerdot{DE}} \\ $$$$=\frac{\left(\frac{{BE}}{{DE}}+\mathrm{1}\right)\centerdot{CE}}{\left(\frac{{AE}}{{CE}}+\mathrm{1}\right)\centerdot{DE}} \\ $$$$=\frac{\frac{{BC}}{{AD}}+\mathrm{1}}{\frac{{BC}}{{AD}}+\mathrm{1}} \\…