Question Number 211716 by liuxinnan last updated on 18/Sep/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\centerdot\centerdot\centerdot×{x}}\right)=? \\ $$ Answered by BHOOPENDRA last updated on 18/Sep/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{n}=\mathrm{0}\:} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{n}!}−\mathrm{1}\right) \\…
Question Number 211717 by liuxinnan last updated on 18/Sep/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{li}{m}}\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}}{{e}^{{x}^{\mathrm{2}} } −\mathrm{1}}=? \\ $$ Answered by BHOOPENDRA last updated on 18/Sep/24 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 211719 by RojaTaniya last updated on 18/Sep/24 Commented by Frix last updated on 18/Sep/24 $${n}=\mathrm{1} \\ $$$$\mathrm{2}+\mathrm{12}+\mathrm{2011}=\mathrm{2025}=\mathrm{45}^{\mathrm{2}} \\ $$ Terms of Service Privacy…
Question Number 211708 by Skyneless last updated on 17/Sep/24 Answered by Frix last updated on 18/Sep/24 $$\int\sqrt{\mathrm{tan}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2tan}^{−\mathrm{1}} \:\sqrt{\mathrm{tan}\:{x}}\right] \\ $$$$=\int\frac{\mathrm{1}−\mathrm{cos}\:{t}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}}{dt} \\ $$$$…
Question Number 211706 by Nadirhashim last updated on 17/Sep/24 $$\:\:\:\:\:\boldsymbol{{if}}\:\:\boldsymbol{{x}}^{\boldsymbol{{log}}\mathrm{27}} +\:\:\:\mathrm{9}^{\boldsymbol{{logx}}} =\mathrm{36}\:\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$ Answered by mehdee7396 last updated on 17/Sep/24 $${x}^{\mathrm{3}{log}\mathrm{3}} +{x}^{\mathrm{2}{log}\mathrm{3}} =\mathrm{36}\:\: \\…
Question Number 211703 by MrGaster last updated on 17/Sep/24 $$ \\ $$$$\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{quadruple}\: \\ $$$$\mathrm{integralas}\:\mathrm{follows}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\boldsymbol{{x}}} \int_{\mathrm{0}} ^{\boldsymbol{\mathrm{y}}} \int_{\mathrm{0}} ^{\boldsymbol{{z}}} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}}…
Question Number 211677 by MrGaster last updated on 16/Sep/24 $$\mathrm{Consider}\:\mathrm{a}\:\mathrm{system}\:\mathrm{consisting}\:\mathrm{ofw} \\ $$$$\mathrm{to}\:\mathrm{masses}.\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{\mathrm{and}}\:\boldsymbol{{m}}_{\mathrm{2}} ,\mathrm{where}\:\mathrm{the}\:\mathrm{pendulum}\:\mathrm{rod}\:\mathrm{has}\:\mathrm{an} \\ $$$$\mathrm{nolinear}\:\mathrm{elastic}\:\mathrm{coefficient}\:\boldsymbol{{k}}\left(\boldsymbol{\theta}\right)=\boldsymbol{{k}}_{\mathrm{0}} \left(\mathrm{1}+\boldsymbol{\alpha\theta}^{\mathrm{2}} \right),\mathrm{with}\boldsymbol{\theta}\:\mathrm{being}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\: \\ $$$$\mathrm{rodand}\:\mathrm{the}\:\mathrm{vertical}\:\mathrm{direction}. \\ $$$$\mathrm{Suppose}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{is}\:\mathrm{smallh} \\ $$$$\mathrm{enoug}\:\mathrm{that}\:\boldsymbol{\theta}\:\mathrm{The}\:\mathrm{square}\:\mathrm{of}\:\mathrm{can}\:\mathrm{be}\:\mathrm{ignored}\: \\…
Question Number 211694 by Tawa11 last updated on 16/Sep/24 Commented by Tawa11 last updated on 16/Sep/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{missing}\:\mathrm{angle} \\ $$ Commented by A5T last updated on…
Question Number 211679 by MrGaster last updated on 16/Sep/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{Inverse}\:\mathrm{root}\:\mathrm{formula}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}=\frac{\mathrm{2}\boldsymbol{{c}}}{−\boldsymbol{{b}}\pm\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{ac}}}} \\ $$$$\left(\mathrm{1}\right)\mathrm{The}“\:\mathrm{antiroot}\:\mathrm{formula}''\:\mathrm{is}\:\mathrm{derivedr} \\ $$$$\mathrm{fom}\:\mathrm{the}\:\mathrm{abovementionedt} \\ $$$$\mathrm{antiroo}\:\mathrm{formula}. \\ $$ Commented by…
Question Number 211673 by a.lgnaoui last updated on 16/Sep/24 $$\:\:\:\boldsymbol{{Resoudre}} \\ $$$$\:\:^{\boldsymbol{\mathrm{x}}} \sqrt{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}−\mathrm{1}}}\:\:\:=\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)^{\left(\boldsymbol{\mathrm{x}}−\mathrm{2}\right)} . \\ $$ Answered by Frix last updated on 16/Sep/24 $$\mathrm{You}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\…