Question Number 103563 by mohammad17 last updated on 15/Jul/20 $${find}\:{laplase}\:{transform}\:{of}\:{the}\:{function} \\ $$$${f}\left({t}\right)={sin}^{\mathrm{2}} {t}\:\:{cos}^{\mathrm{3}} {t}\:\:? \\ $$ Answered by MAB last updated on 15/Jul/20 $${f}\left({t}\right)=\left(\mathrm{1}−{cos}^{\mathrm{2}} {t}\right){cos}^{\mathrm{3}}…
Question Number 38025 by mondodotto@gmail.com last updated on 20/Jun/18 $$\boldsymbol{\mathrm{An}}\:\boldsymbol{\mathrm{AP}}\:\boldsymbol{\mathrm{has}}\:\mathrm{41}\:\boldsymbol{\mathrm{terms}}.\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{first}}\:\boldsymbol{\mathrm{five}} \\ $$$$\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{AP}}\:\boldsymbol{\mathrm{is}}\:\mathrm{35}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{last}}\:\boldsymbol{\mathrm{five}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{AP}}\:\boldsymbol{\mathrm{is}}\:\mathrm{395}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{common}}\:\boldsymbol{\mathrm{difference}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{first}}\:\boldsymbol{\mathrm{term}}. \\ $$ Answered by ajfour last updated on 20/Jun/18…
Question Number 169044 by mokys last updated on 23/Apr/22 $${does}\:{the}\:{series}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}\left({n}−\mathrm{1}\right)} \:{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)\:{is}\:{converge}\:{or}\:{diverge}\:? \\ $$ Answered by Mathspace last updated on 23/Apr/22 $${n}=\mathrm{2}{m}\Rightarrow{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${n}=\mathrm{2}{m}+\mathrm{1}\Rightarrow{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)={sin}\left(\frac{\left(\mathrm{2}{m}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)…
Question Number 37964 by naka3546 last updated on 20/Jun/18 Answered by gunawan last updated on 20/Jun/18 $$\mathrm{put}\: \\ $$$$\left({a}+{b}\right)\geqslant\mathrm{2}{c} \\ $$$$\frac{\left({a}+{b}\right)\left({a}−{b}\right)}{{c}}\geqslant\mathrm{2}\left({a}−{b}\right) \\ $$$$\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{c}}\geqslant\mathrm{2}{a}−\mathrm{2}{b}\:\:\:..\left({i}\right)…
Question Number 103492 by abony1303 last updated on 15/Jul/20 $$\mathrm{Q1}:\:\:\mathrm{Evaluate}\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \mid\mathrm{x}\mid\centerdot\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{Q2}:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{integers}\:{k}\:\mathrm{for}\: \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +{k}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one}\:\mathrm{solution}. \\ $$$$…
Question Number 103489 by Study last updated on 15/Jul/20 Answered by bobhans last updated on 15/Jul/20 $${AB}=\:\sqrt{\mathrm{6}}\:+\:\mathrm{2}\sqrt{\mathrm{6}}\:=\:\mathrm{3}\sqrt{\mathrm{6}}\:\oplus \\ $$ Answered by bramlex last updated on…
Question Number 103480 by mohammad17 last updated on 15/Jul/20 $${if}\:{f}\left({x}\right)=\mid{x}−\mathrm{1}\:\:\:\:\:\:\mathrm{0}<{x}<\mathrm{1}\mid\:{solve}\:{in} \\ $$$$ \\ $$$$\left(\mathrm{1}\right){Fourier}\:{series}\:{of}\:{sines}\:{only}\:? \\ $$$$\left(\mathrm{2}\right)\:{Fourier}\:{series}\:{of}\:{cosines}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103481 by mohammad17 last updated on 15/Jul/20 $${prove}\:{that}\:\left({cos}\frac{{n}\theta}{\mathrm{5}}+{isin}\frac{\mathrm{2}{n}\theta}{\mathrm{10}}\right)^{\mathrm{5}} −\frac{\mathrm{1}}{{e}^{{in}\theta} }=\mathrm{2}{sin}\left({n}\theta\right)\:? \\ $$ Answered by Dwaipayan Shikari last updated on 15/Jul/20 $${e}^{{n}\theta{i}} −{e}^{−{in}\theta} =\mathrm{2}{sin}\left({n}\theta\right)…
Question Number 103478 by mohammad17 last updated on 15/Jul/20 Answered by bemath last updated on 15/Jul/20 $${HE}\:\Rightarrow\lambda^{\mathrm{2}} +\mathrm{3}\lambda−\mathrm{2}=\mathrm{0} \\ $$$$\left(\lambda+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\lambda+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{17}}{\mathrm{4}}\:\Rightarrow\lambda=−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{17}}}{\mathrm{2}} \\…
Question Number 169013 by 0731619 last updated on 23/Apr/22 Commented by cortano1 last updated on 23/Apr/22 $$\:=\:\sqrt{\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}}{{x}+\sqrt{{x}+\sqrt{{x}}}}}\: \\ $$$$=\:\sqrt{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\frac{\mathrm{1}}{{x}}+\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}}}} \\ $$$$=\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}\:=\:\mathrm{1} \\…