Question Number 143134 by help last updated on 10/Jun/21 Answered by liberty last updated on 10/Jun/21 $$\:{y}'=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\left({x}+{h}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{h}\right)\right)−\mathrm{cos}\:\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left({x}^{\mathrm{2}} +\mathrm{2}{hx}+{h}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}{h}\right)−\mathrm{cos}\:\left({x}^{\mathrm{2}}…
Question Number 12023 by tawa last updated on 09/Apr/17 $$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:−\:\sqrt{\mathrm{xy}}\:=\:\mathrm{3}\:\:\:\:\:……….\:\left(\mathrm{i}\right) \\ $$$$\sqrt{\mathrm{x}\:+\:\mathrm{1}}\:+\:\sqrt{\mathrm{y}\:+\:\mathrm{1}}\:=\:\mathrm{4}\:\:\:\:\:……….\:\left(\mathrm{ii}\right) \\ $$ Answered by Mr Chheang Chantria last updated on 10/Apr/17…
Question Number 11994 by algabara last updated on 09/Apr/17 $$\mathrm{4}{x}=\mathrm{60} \\ $$ Commented by tawa last updated on 09/Apr/17 $$\mathrm{4x}\:\:=\:\mathrm{60} \\ $$$$\mathrm{Divide}\:\mathrm{both}\:\mathrm{side}\:\mathrm{by}\:\mathrm{4} \\ $$$$\frac{\mathrm{4x}}{\mathrm{4}}\:=\:\frac{\mathrm{60}}{\mathrm{4}} \\…
Question Number 142902 by greg_ed last updated on 07/Jun/21 $$\mathrm{I}_{{n}} =\int_{\mathrm{0}} ^{\:_{} \frac{\pi}{\mathrm{2}}} \:\left(\mathrm{sin}\:{x}\right)^{{n}} \:{dx} \\ $$$$\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{integration}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{parts}},\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\::\: \\ $$$$\mathrm{I}_{{n}+\mathrm{2}} \:=\:\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\:.\:\mathrm{I}_{{n}} \\ $$ Answered by qaz…
Question Number 11823 by Joel576 last updated on 01/Apr/17 $$\mathrm{3}\:+\:\mathrm{2}\left(\mathrm{3}^{\mathrm{2}} \right)\:+\:\mathrm{3}\left(\mathrm{3}^{\mathrm{3}} \right)\:+\:…\:+\:\mathrm{10}\left(\mathrm{3}^{\mathrm{10}} \right)\:=\:? \\ $$ Answered by ajfour last updated on 02/Apr/17 $${S}=\:\mathrm{3}+\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{3}\right)^{\mathrm{3}} +…+\mathrm{10}\left(\mathrm{3}\right)^{\mathrm{10}}…
Question Number 77314 by aliesam last updated on 05/Jan/20 Commented by mind is power last updated on 05/Jan/20 $$\mathrm{first}\:\Psi\left(\frac{−\mathrm{2}}{\varphi}\right)\mathrm{is}\:\mathrm{psi}\:\mathrm{function}? \\ $$ Commented by aliesam last…
Question Number 77285 by jagoll last updated on 05/Jan/20 $$\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\mathrm{Fourier}\:\mathrm{series}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}\:,\:\mathrm{0}\:<\:\mathrm{x}<\frac{\mathrm{1}}{\mathrm{8}} \\ $$ Answered by john santu last updated on 05/Jan/20 $$\mathrm{because}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function}\: \\ $$$$,\:\mathrm{we}\:\mathrm{used}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{series}.…
Question Number 11743 by saa last updated on 30/Mar/17 $$\mathrm{9}/\mathrm{3}\left(\mathrm{6}×\mathrm{4}/\mathrm{8}\right)=? \\ $$ Answered by Joel576 last updated on 31/Mar/17 $$\frac{\mathrm{9}}{\mathrm{3}\left(\mathrm{6}\:.\:\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{9}}{\mathrm{3}\:.\:\mathrm{3}}\:=\:\mathrm{1} \\ $$ Terms of Service…
Question Number 11732 by agni5 last updated on 30/Mar/17 $$\mathrm{0}.\mathrm{16}\:\boldsymbol{\div}\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{of}\:\frac{\mathrm{2}}{\mathrm{5}}\:\boldsymbol{\div}\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$ Answered by ajfour last updated on 30/Mar/17 $$\mathrm{8} \\ $$ Terms of Service…
Question Number 77132 by peter frank last updated on 03/Jan/20 Commented by kaivan.ahmadi last updated on 03/Jan/20 $${z}=\mathrm{4}\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{4}\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{6}{i}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}{i}=\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{4}{i}=\mathrm{4}\left(\sqrt{\mathrm{3}}+{i}\right)=\mathrm{8}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{8}{e}^{{i}\frac{\pi}{\mathrm{6}}} \\ $$$$\Rightarrow\frac{{z}}{\mathrm{8}}+{i}\left(\frac{{z}}{\mathrm{8}}\right)^{\mathrm{2}} +\left(\frac{{z}}{\mathrm{8}}\right)^{\mathrm{3}}…