Question Number 208681 by Noorzai last updated on 20/Jun/24 Answered by Rasheed.Sindhi last updated on 21/Jun/24 $$\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\:=\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}\:={a}+{b}\sqrt{\mathrm{3}}\:\left(>\mathrm{0}\right)\left({let}\right) \\ $$$$\:{Where}\:{a},{b}\in\mathbb{Z} \\ $$$$\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:={a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{\mathrm{3}}\: \\ $$$${a}^{\mathrm{2}}…
Question Number 208676 by RoseAli last updated on 20/Jun/24 Answered by Berbere last updated on 20/Jun/24 $$\sqrt[{\mathrm{4}}]{{a}}+\sqrt[{\mathrm{4}}]{{b}}=\sqrt{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}=\sqrt[{\mathrm{4}}]{\mathrm{2}}.\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{2}{a}}+\sqrt[{\mathrm{4}}]{\mathrm{2}{b}}=\sqrt{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}+\sqrt[{\mathrm{4}}]{\frac{{b}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$${b}=\mathrm{0}\:\Rightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1}\:{no}\:{solution}\:\mathbb{N} \\…
Question Number 208645 by Mastermind last updated on 20/Jun/24 $$\mathrm{Solve}\:: \\ $$$$\mathrm{2x}_{\mathrm{1}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{5}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{2}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{2}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{3}} \:−\:\mathrm{3}\lambda_{\mathrm{1}} \:−\:\lambda_{\mathrm{2}} \:=\:\mathrm{0}…
Question Number 208646 by hardmath last updated on 20/Jun/24 $$\mathrm{y}\:=\:\mid\mathrm{x}\:−\:\mathrm{2}\mid\:+\:\mid\mathrm{x}\:+\:\mathrm{4}\mid \\ $$$$\mathrm{Find}:\:\:\:\mathrm{min}\left(\mathrm{y}\right)\:\:\:\mathrm{and}\:\:\:\mathrm{max}\left(\mathrm{y}\right) \\ $$ Answered by A5T last updated on 20/Jun/24 $${y}\:{has}\:{no}\:{global}\:{maximum},{y}\rightarrow+\infty\:{as}\:{x}\rightarrow\underset{−} {+}\infty \\ $$$$\mid{x}−\mathrm{2}\mid+\mid−\mathrm{4}−{x}\mid\geqslant\mid{x}−\mathrm{2}−\mathrm{4}−{x}\mid=\mathrm{6}\Rightarrow{min}\left({y}\right)=\mathrm{6}…
Question Number 208647 by hardmath last updated on 20/Jun/24 $$\left(\mathrm{tan}^{\mathrm{2}} \boldsymbol{\mathrm{x}}\:−\:\mathrm{3}\right)\:\centerdot\:\mathrm{sin}\boldsymbol{\mathrm{x}}\:=\:\mathrm{0} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Frix last updated on 20/Jun/24 $${f}\left({x}\right)×{g}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:{f}\left({x}\right)=\mathrm{0}\vee{g}\left({x}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\:=\mathrm{0}\:\Rightarrow\:{x}={n}\pi…
Question Number 208623 by MWSuSon last updated on 19/Jun/24 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x},\:\mathrm{3}^{\mathrm{x}} −\mathrm{2}^{\mathrm{x}} =\mathrm{65} \\ $$ Answered by Berbere last updated on 19/Jun/24 $${if}\:{x}<\mathrm{0} \\ $$$$\mathrm{3}^{{x}} <\mathrm{1\&2}^{{x}}…
Question Number 208632 by necx122 last updated on 19/Jun/24 $$\int{e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$${could}\:{this}\:{be}\:{integrated}\:{by}\:{part}?\:{What} \\ $$$${approach}\:{would}\:{most}\:{likely}\:{be}\:{suitable} \\ $$$${for}\:{this}\:{integral}? \\ $$$$ \\ $$ Commented by Tinku…
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Question Number 208624 by vipin last updated on 19/Jun/24 Answered by Berbere last updated on 19/Jun/24 $$\frac{\mathrm{1}}{{cosec}^{−} \left(−\sqrt{\mathrm{2}}\right)}=\mathrm{sin}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=−\frac{\pi}{\mathrm{4}} \\ $$$${f}\left({x}\right).\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{\mathrm{1}+\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}\right)…{E} \\ $$$${g}\left({y}\right)=\mathrm{tan}^{−\mathrm{1}}…