Question Number 109584 by bobhans last updated on 24/Aug/20 $$\:\:\:\frac{−\flat{o}\flat−}{{hans}} \\ $$$$\left(\mathrm{1}\right)\sqrt{{x}−\sqrt{{x}−\frac{\mathrm{1}}{\mathrm{4}}}}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\mid\overset{\rightarrow} {{a}}\mid\:=\:\mathrm{1},\:\mid\overset{\rightarrow} {{b}}\mid\:=\:\mathrm{2}\:,\:\mid\overset{\rightarrow} {{c}}\mid=\mathrm{3}\:,\:\angle\left(\overset{\rightarrow} {{a}},\overset{\rightarrow} {{b}}\right)=\mathrm{90}° \\ $$$$\:\:\:\:\:\:\:\:\angle\left(\overset{\rightarrow} {{b}},\overset{\rightarrow} {{c}}\right)=\mathrm{60}°\:,\:\angle\left(\overset{\rightarrow} {{a}},\overset{\rightarrow} {{c}}\right)=\mathrm{120}°\:,\:{then}\:…
Question Number 109574 by mathdave last updated on 24/Aug/20 Commented by mathdave last updated on 24/Aug/20 $${solution}\:{to}\:{earlier}\:{post} \\ $$ Commented by Sarah85 last updated on…
Question Number 175095 by Gbenga last updated on 18/Aug/22 $$\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{c}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$ Answered by MJS_new last updated on 18/Aug/22 $${x}={z}−\frac{\mathrm{1}}{\mathrm{3}} \\…
Question Number 109553 by mathdave last updated on 24/Aug/20 Commented by ajfour last updated on 24/Aug/20 $${why}? \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 175085 by MathsFan last updated on 18/Aug/22 Commented by MathsFan last updated on 18/Aug/22 $$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{help}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 175061 by daus last updated on 17/Aug/22 Commented by Rasheed.Sindhi last updated on 17/Aug/22 $$\left(\mathrm{a}\right)\:\:\mathrm{150} \\ $$ Answered by cortano1 last updated on…
Question Number 109497 by rinasitorus_ last updated on 24/Aug/20 Answered by nurmaya_silaban last updated on 24/Aug/20 $$\left.\mathrm{1}\right){a}.\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{5}}=\:\frac{\mathrm{5}}{\mathrm{15}}+\frac{\mathrm{9}}{\mathrm{15}}=\frac{\mathrm{14}}{\mathrm{15}} \\ $$$$\:\:\:\:\:\mathrm{b}.\:\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{3}}{\mathrm{15}} \\ $$$$\:\:\:\:\:\mathrm{c}.\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{7}}=\:\frac{\mathrm{14}}{\mathrm{21}}−\frac{\mathrm{9}}{\mathrm{21}}=\:\frac{\mathrm{5}}{\mathrm{21}} \\ $$$$\:\:\:\:\:\mathrm{d}.\frac{\mathrm{2}}{\mathrm{3}}:\frac{\mathrm{3}}{\mathrm{7}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{7}}{\mathrm{3}}=\frac{\mathrm{14}}{\mathrm{9}} \\ $$$$…
Question Number 175028 by sciencestudent last updated on 16/Aug/22 Commented by mr W last updated on 17/Aug/22 $$\left({a}\right) \\ $$$${R}=\mathrm{990}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{750}}+\frac{\mathrm{1}}{\mathrm{680}}}=\mathrm{1346}.\mathrm{6}\Omega \\ $$ Terms of Service…
Question Number 109494 by mathdave last updated on 24/Aug/20 Answered by mathmax by abdo last updated on 24/Aug/20 $$\mathrm{for}\:\mid\mathrm{u}\mid<\mathrm{1}\:\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{d}}{\mathrm{du}}\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}} \:\Rightarrow \\ $$$$\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)\:=\sum_{\mathrm{n}=\mathrm{0}}…
Question Number 109489 by bobhans last updated on 24/Aug/20 $$\left(\mathrm{1}\right)\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\left({x}\right)+\mathrm{cos}\:\left({x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{sin}\:{x}} \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\mid{x}+\mathrm{1}\mid\:\mathrm{sin}\:\left({x}+\mathrm{1}\right) \\ $$$$ \\ $$ Answered by john…