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Question Number 119588 by zakirullah last updated on 25/Oct/20 Answered by bemath last updated on 25/Oct/20 $$\left({i}\right)\:{let}\:\overset{\rightarrow} {{c}}\:{is}\:{unit}\:{vector}\:{such}\:{that} \\ $$$${orthogonal}\:{to}\:\overset{\rightarrow} {{a}}\:\&\:\overset{\rightarrow} {{b}} \\ $$$$\:{so}\:{gives}\:\overset{\rightarrow} {{c}}\:=\:\pm\:\frac{\mid\overset{\rightarrow}…
Question Number 119572 by zakirullah last updated on 25/Oct/20 Commented by bemath last updated on 25/Oct/20 $$\left({i}\right)\:{method}\:\mathrm{1} \\ $$$$\:\overset{\rightarrow} {{a}}\:\parallel\:\overset{\rightarrow} {{b}}\:{if}\:\overset{\rightarrow} {{a}}\:×\:\overset{\rightarrow} {{b}}\:=\:\overset{\rightarrow} {\mathrm{0}} \\…
Question Number 185085 by Ml last updated on 16/Jan/23 $$\mathrm{log}_{\mathrm{3}} \left(\mathrm{a}+\mathrm{1}\right)=\mathrm{log}_{\mathrm{4}} \left(\mathrm{a}+\mathrm{8}\right) \\ $$$$\mathrm{a}=?? \\ $$$$\mathrm{please}\:\mathrm{solution} \\ $$ Answered by Frix last updated on 16/Jan/23…
Question Number 185090 by SEKRET last updated on 16/Jan/23 $$\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1}\:=\:\:\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=? \\ $$$$ \\ $$$$ \\ $$…
Question Number 185087 by SEKRET last updated on 16/Jan/23 $$ \\ $$$$ \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2004}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}\in\boldsymbol{{Z}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}=?\:\:\: \\ $$$$ \\…
Question Number 185086 by SEKRET last updated on 16/Jan/23 $$ \\ $$$$\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\boldsymbol{\mathrm{n}}} }\:=\:? \\ $$$$…
Question Number 185075 by aba last updated on 16/Jan/23 $$ \\ $$$$\mathrm{2}^{\mathrm{x}} =\mathrm{4x}\:\Rightarrow\:\:\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{x}} }=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}.\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow−\mathrm{xln}\left(\mathrm{2}\right)\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2x}\right)} =−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{W}\left(−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)}…
Question Number 119524 by help last updated on 25/Oct/20 Answered by benjo_mathlover last updated on 25/Oct/20 $${g}\left({x}\right)\:=\:\sqrt{\left(−\mathrm{1}\right)\left({x}−\mathrm{14}\right)}\:=\:\sqrt{{x}−\mathrm{14}}\:{i}\: \\ $$$${thus}\:{f}\left({x}\right)=\sqrt{{x}−\mathrm{14}}\:+\:{i}\:\sqrt{{x}−\mathrm{14}}\:=\:\left({i}+\mathrm{1}\right)\sqrt{{x}−\mathrm{14}} \\ $$ Terms of Service Privacy…
Question Number 185024 by aba last updated on 15/Jan/23 $$\mathrm{prove}\:\mathrm{that}\::\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}<\mathrm{ln}\left(\mathrm{2}\right) \\ $$ Answered by witcher3 last updated on 16/Jan/23 $${U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}+{k}},{U}_{{n}+\mathrm{1}}…