Question Number 114485 by mohammad17 last updated on 19/Sep/20 Answered by Dwaipayan Shikari last updated on 19/Sep/20 $${f}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}}=\frac{\mathrm{1}}{{x}\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)} \\ $$$${x}\in{R}−\left\{\mathrm{0},\mathrm{1},\mathrm{2}\right\}\:\:\left({Domain}\right) \\ $$ Commented…
Question Number 114482 by mohammad17 last updated on 19/Sep/20 Answered by 1549442205PVT last updated on 19/Sep/20 $$\left.\mathrm{Choose}\:\mathrm{ii}\right)−\mathrm{2},\mathrm{3},\mathrm{1}\:\mathrm{since} \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{x}\in\left\{−\mathrm{2};\mathrm{1};\mathrm{3}\right\}…
Question Number 114480 by mohammad17 last updated on 19/Sep/20 Commented by mohammad17 last updated on 19/Sep/20 $${help}\:{me}\:{sir} \\ $$ Answered by john santu last updated…
Question Number 114475 by Rio Michael last updated on 19/Sep/20 Commented by Rio Michael last updated on 19/Sep/20 $$\mathrm{suppose}\:\mathrm{that}\:\mathrm{two}\:\mathrm{rocks}\:\mathrm{are}\:\mathrm{thrown}\:\mathrm{from} \\ $$$$\mathrm{thesame}\:\mathrm{point}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{moment}\:\mathrm{as}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{figure}\:\mathrm{above}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{them}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}.\:\mathrm{Assume}\:\mathrm{that}…
Question Number 179999 by yaslm last updated on 05/Nov/22 Commented by mokys last updated on 05/Nov/22 $${lnx}\:−\:\mathrm{1}\:=\:{y}\:\rightarrow\:{lnx}\:=\:{y}\:+\:\mathrm{1}\rightarrow\:{x}\:=\:{e}^{{y}+\mathrm{1}} \rightarrow\:{dx}\:=\:{e}^{{y}+\mathrm{1}} {dy} \\ $$$$ \\ $$$${I}\:=\:\int\:\left({y}+\mathrm{1}\right)\:{e}^{{y}^{\mathrm{2}} +{y}} \:{e}^{{y}+\mathrm{1}}…
Question Number 48916 by Tinku Tara last updated on 30/Nov/18 $$\mathrm{Dear}\:\mathrm{Mr}\:\mathrm{W2} \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{send}\:\mathrm{us}\:\mathrm{an}\:\mathrm{email}?\:\mathrm{We} \\ $$$$\mathrm{need}\:\mathrm{to}\:\mathrm{figure}\:\mathrm{out}\:\mathrm{how}\:\mathrm{to}\:\mathrm{troubleshoot} \\ $$$$\mathrm{problem}\:\mathrm{seen}\:\mathrm{on}\:\mathrm{the}\:\mathrm{device}\:\mathrm{you} \\ $$$$\mathrm{mentioned}. \\ $$ Commented by…
Question Number 114443 by ARVIND990 last updated on 19/Sep/20 Commented by ARVIND990 last updated on 19/Sep/20 $$\mathrm{fine}\:\mathrm{the}\:\mathrm{projection}\:\mathrm{of}\:\mathrm{point}\:\overset{\rightarrow} {\mathrm{OP}}=\overset{\rightarrow} {\mathrm{p}}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{line}\:\overset{\rightarrow} {\mathrm{r}}=\overset{\rightarrow} {\mathrm{a}}+\lambda\overset{\rightarrow} {\mathrm{b}} \\…
Question Number 48899 by behi83417@gmail.com last updated on 29/Nov/18 Commented by behi83417@gmail.com last updated on 29/Nov/18 $${ABC},{equilateral}. \\ $$$${CD}=\mathrm{8},{CE}=\mathrm{10},{AD}=\mathrm{5},{DE}=\mathrm{4},{EB}=\mathrm{3} \\ $$$$\Rightarrow{AB}=? \\ $$ Answered by…
Question Number 48869 by behi83417@gmail.com last updated on 29/Nov/18 Commented by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18 $${i}\:{can}\:{not}\:{understand}\:{the}\:{questions}.. \\ $$$${let}\:\frac{{P}}{{Q}}={Intregal}\:{part}+{fractional}\:{part} \\ $$$${in}\:{which}\:{location}\:{zeros}\:{to}\:{find}.. \\ $$$$ \\ $$$$…
Question Number 179936 by yaslm last updated on 04/Nov/22 Answered by Frix last updated on 04/Nov/22 $${i} \\ $$$${f}\left({x}\right)=\begin{cases}{{x}\left({x}−\mathrm{2}\right);\:{x}\leqslant\mathrm{0}}\\{−{x}\left({x}−\mathrm{2}\right);\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}}\\{{x}\left({x}−\mathrm{2}\right);\:{x}\geqslant\mathrm{2}}\end{cases} \\ $$$${f}'\left({x}\right)=\begin{cases}{\mathrm{2}{x}−\mathrm{2};\:{x}\leqslant\mathrm{0}}\\{−\mathrm{2}{x}+\mathrm{2};\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}}\\{\mathrm{2}{x}−\mathrm{2};\:{x}\geqslant\mathrm{2}}\end{cases} \\ $$$${f}'\left(\mathrm{0}\right)=−\mathrm{2}\wedge{f}'\left(\mathrm{0}\right)=\mathrm{2} \\ $$$${f}'\left(\mathrm{2}\right)=−\mathrm{2}\wedge{f}'\left(\mathrm{2}\right)=\mathrm{2}…