Question Number 195206 by otchereabdullai@gmail.com last updated on 27/Jul/23 Answered by MM42 last updated on 27/Jul/23 $${p}\left({a}\right)={p}\left({b}\right)={p}\left({c}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:;\:{probability}\:{of}\:{winning}\:{each}\:{race} \\ $$$$\left({i}\right)\:{p}\left({a}'\cap{b}\cap{c}'\right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{64}} \\ $$$$\left({ii}\right)\:{p}\left({a}\cap{b}\cap{c}\right)=\frac{\mathrm{27}}{\mathrm{64}} \\ $$$$\left({iii}\right)\:{p}\left(\Sigma{abc}'\right)=\mathrm{3}×\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{9}}{\mathrm{16}}=\frac{\mathrm{27}}{\mathrm{64}} \\ $$…
Question Number 195203 by Erico last updated on 27/Jul/23 $$\mathrm{Calculer}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \frac{\mathrm{dt}}{\left(\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} \right)^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} } \\ $$ Answered by Frix last updated on 27/Jul/23…
Question Number 195202 by sonukgindia last updated on 27/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
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}}…