Question Number 207904 by Ahmed777hamouda last updated on 30/May/24 Commented by Ahmed777hamouda last updated on 30/May/24 $$\mathrm{H}{ow}\:\boldsymbol{\mathrm{prove}}\:\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} =\mathrm{2}\boldsymbol{\mathrm{I}}_{{o}} \boldsymbol{\mathrm{J}}_{{n}} \left(\boldsymbol{\mathrm{nx}}\right)\mathrm{cos}\:\left(\boldsymbol{\mathrm{n}}\theta_{\boldsymbol{\mathrm{g}}} +\boldsymbol{\mathrm{n}\theta}_{\boldsymbol{\mathrm{n}}} \right) \\ $$$$\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{fourier}}\:\boldsymbol{\mathrm{series}} \\…
Question Number 207906 by nachosam last updated on 30/May/24 $${help} \\ $$$$\int_{\mathrm{1}} ^{\:\infty} {x}^{−{ln}\left({x}\right)} {dx} \\ $$$$ \\ $$ Answered by Berbere last updated on…
Question Number 207901 by Tawa11 last updated on 29/May/24 Answered by mr W last updated on 30/May/24 $$\boldsymbol{{v}}=\frac{{d}\boldsymbol{{r}}}{{dt}}={e}^{{t}} \left(\mathrm{cos}\:{t}−\mathrm{sin}\:{t}\right)\:\boldsymbol{{i}}+{e}^{{t}} \left(\mathrm{sin}\:{t}+\:\mathrm{cos}\:{t}\right)\:\boldsymbol{{j}} \\ $$$$\boldsymbol{{a}}=\frac{{d}\boldsymbol{{v}}}{{dt}}=−\mathrm{2}{e}^{{t}} \:\mathrm{sin}\:{t}\:\boldsymbol{{i}}+\mathrm{2}{e}^{{t}} \:\mathrm{cos}\:{t}\:\boldsymbol{{j}} \\…
Question Number 207885 by hardmath last updated on 29/May/24 $$\mathrm{Find}: \\ $$$$\boldsymbol{\mathrm{i}}^{\mathrm{4}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{8}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{12}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{16}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{20}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{24}} \:+…+\:\boldsymbol{\mathrm{i}}^{\mathrm{100}} \:=\:? \\ $$ Commented by mr W…
Question Number 207864 by efronzo1 last updated on 29/May/24 $$\:\:\:\:\:\underbrace{\:} \\ $$ Answered by Rasheed.Sindhi last updated on 29/May/24 $$\mathrm{2}^{\mathrm{5}{m}} \centerdot\mathrm{5}^{\mathrm{2}{n}} \centerdot{k}=\mathrm{2020}^{\mathrm{2020}} =\left(\mathrm{2}^{\mathrm{2}} .\mathrm{5}.\mathrm{101}\right)^{\mathrm{2020}} \\…
Question Number 207897 by hardmath last updated on 29/May/24 $$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{12}\:\:\centerdot\:\:\mathrm{13}\:\:\centerdot\:\:\mathrm{14}\:\:\centerdot\:\:\mathrm{15}\:\:+\:\:\mathrm{1}}\:\:=\:\:? \\ $$ Answered by Frix last updated on 29/May/24 $$\sqrt{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)+\mathrm{1}}= \\ $$$$=\sqrt{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}}…
Question Number 207866 by efronzo1 last updated on 29/May/24 $$\:\:\:\mathrm{If}\:\mathrm{the}\:\mathrm{system}\:\begin{cases}{\mathrm{y}=−\mathrm{mx}^{\mathrm{2}} −\mathrm{2}}\\{\mathrm{4x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\:\mathrm{4}}\end{cases} \\ $$$$\:\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\: \\ $$$$\:\:\mathrm{them}\:\mathrm{m}\:=\: \\ $$$$\:\:\left(\mathrm{A}\right)\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\left(\mathrm{B}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\:\left(\mathrm{C}\right)\:\mathrm{1}\:\:\:\:\left(\mathrm{D}\right)\:\sqrt{\mathrm{2}}\:\:\:\left(\mathrm{E}\right)\:\sqrt{\mathrm{3}} \\ $$$$\:\: \\ $$ Answered by…
Question Number 207876 by hardmath last updated on 29/May/24 $$\mathrm{Find}:\:\:\:\mathrm{ln}\:\left(\frac{\mathrm{2}\:\mathrm{tg}\:\mathrm{22},\mathrm{30}°}{\mathrm{1}\:−\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{22},\mathrm{30}°}\right)\:=\:? \\ $$ Commented by mr W last updated on 29/May/24 $${use}\:{your}\:{calculator}! \\ $$ Answered…
Question Number 207878 by luciferit last updated on 29/May/24 Answered by Berbere last updated on 30/May/24 $${not}\:{well}\:{defind} \\ $$$$\left.{g}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {f}\left({t}^{\mathrm{4}} \right)+\mathrm{4}{t}^{\mathrm{4}} {f}'\left({t}\right)\right){dt}? \\ $$$$…
Question Number 207879 by efronzo1 last updated on 29/May/24 Answered by mr W last updated on 29/May/24 $${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{12}=\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}\geqslant\mathrm{3} \\ $$$${f}\left({a}\right)=\mathrm{65539}={a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{12} \\ $$$$\Rightarrow{a}^{\mathrm{2}}…