Question Number 195180 by mustafazaheen last updated on 26/Jul/23 $$ \\ $$$$\mathrm{1}.\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{sinx}\:\:\:\:\:\:,\:\:\:\:\frac{\pi}{\mathrm{2}}<\mathrm{x}\leqslant\mathrm{2}\pi}\\{\mathrm{cosx}\:\:\:\:\:\:,\:\:\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{f}^{'} \left(\frac{\pi}{\mathrm{2}}\right)\:=? \\ $$$$\mathrm{2}.\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{sinx}\:\:\:\:\:\:,\:\:\:\:\frac{\pi}{\mathrm{2}}<\mathrm{x}\leqslant\mathrm{2}\pi}\\{\mathrm{cosx}\:\:\:\:\:\:,\:\:\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{f}'\left(\mathrm{2}\pi\right)\:=? \\ $$$$ \\ $$ Answered by…
Question Number 195192 by mustafazaheen last updated on 26/Jul/23 $$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{sinx}+\frac{\pi}{\mathrm{2}}\right)^{\mathrm{x}} =? \\ $$ Answered by MM42 last updated on 26/Jul/23 $${sinx}+\frac{\pi}{\mathrm{2}}>\mathrm{1} \\…
Question Number 195195 by Mr.D.N. last updated on 26/Jul/23 $$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{limit}: \\ $$$$\:\underset{\mathrm{x}\rightarrow\mathrm{a}\:} {\overset{\mathrm{lim}} {\:}}\:\:\:\frac{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}}} }\:−\:\frac{\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$ Commented by Frix last updated…
Question Number 195194 by Abdullahrussell last updated on 26/Jul/23 Answered by Frix last updated on 26/Jul/23 $${a}\:\:\:\:\:{b}\:\:\:\:\:{c}\:\:\:\:\:{d} \\ $$$$\mathrm{2}\:\:\:\:\:\mathrm{3}\:\:\:\:\mathrm{15}\:\:\:\mathrm{10} \\ $$$$\mathrm{2}\:\:\:\:\:\mathrm{4}\:\:\:\:\mathrm{12}\:\:\:\:\mathrm{6} \\ $$$$\mathrm{2}\:\:\:\:\:\mathrm{6}\:\:\:\:\mathrm{12}\:\:\:\:\mathrm{4} \\ $$$$\mathrm{2}\:\:\:\:\mathrm{10}\:\:\:\mathrm{15}\:\:\:\:\mathrm{3}…
Question Number 195178 by Shlock last updated on 26/Jul/23 Answered by mr W last updated on 26/Jul/23 $${DC}={BC}=\mathrm{2},\:{say} \\ $$$${KD}=\mathrm{1} \\ $$$$\angle{KCD}=\alpha,\:{say} \\ $$$$\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\mathrm{60}°\right)\:\mathrm{tan}\:\left(\mathrm{60}°−\alpha\right)=\mathrm{2}\:\mathrm{sin}\:\mathrm{60}° \\…
Question Number 195191 by Erico last updated on 26/Jul/23 $$\mathrm{Calculer}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\:\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{n}} }\right)−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\mathrm{sinx}\:\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{n}} }\right)}{\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{n}} }\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195185 by Erico last updated on 26/Jul/23 $$\mathrm{1}/\:\:\mathrm{Montrer}\:\mathrm{que}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \:\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{p}−\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{4}{p}} }{dx}=\left(\frac{\mathrm{2}^{\mathrm{2}{p}−\mathrm{3}} }{{p}}\right)\pi\left[\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{p}−\mathrm{1}} {\sum}}{cos}^{\mathrm{2}{p}−\mathrm{1}} \left(\frac{{k}\pi}{\mathrm{2}{p}}\right)\right] \\ $$$$\mathrm{2}/\:\:\:\:\mathrm{En}\:\mathrm{d}\acute {\mathrm{e}duire}\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{p}−\mathrm{1}}…
Question Number 195165 by Nimnim111118 last updated on 25/Jul/23 $$\mathrm{If}\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{1}} }=\left(\mathrm{sin}\theta,\mathrm{cos}\theta,\theta\right),\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{2}} }=\left(\mathrm{cos}\theta,−\mathrm{sin}\theta,−\mathrm{3}\right)\:\mathrm{and} \\ $$$$\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{3}} }=\left(\mathrm{2},\mathrm{3},−\mathrm{1}\right),\:\mathrm{find}\:\frac{\mathrm{d}}{\mathrm{d}\theta}\left\{\overset{\rightarrow} {\mathrm{r}_{\mathrm{1}} }×\left(\overset{\rightarrow} {\mathrm{r}_{\mathrm{2}} }×\overset{\rightarrow} {\mathrm{r}_{\mathrm{3}} }\right)\right\}\:\mathrm{at}\:\theta=\mathrm{0} \\…
Question Number 195135 by 073 last updated on 25/Jul/23 Answered by som(math1967) last updated on 25/Jul/23 $$\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}{x}.\frac{\mathrm{1}}{{x}}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\:\frac{{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 195129 by a.lgnaoui last updated on 25/Jul/23 $$\mathrm{Calculer}\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{de}\:\mathrm{la}\:\mathrm{serie}\:\mathrm{suivante}: \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{7}}{\mathrm{32}}+\frac{\mathrm{9}}{\mathrm{128}}+….. \\ $$ Answered by MM42 last updated on 25/Jul/23 $$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }{S}=\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }+\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}}…