Question Number 42357 by preet last updated on 24/Aug/18 Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18 $$\overset{\rightarrow} {{v}}=\frac{{d}\overset{\rightarrow} {{r}}}{{dt}}=\frac{{d}}{{dt}}\left(\mathrm{3}{ti}−{t}^{\mathrm{2}} {j}+\mathrm{4}{k}\right) \\ $$$$\overset{\rightarrow} {{v}}=\mathrm{3}{i}−\mathrm{2}{tj}+\mathrm{0}.{k} \\ $$$$\left(\overset{\rightarrow}…
Question Number 42196 by rahul 19 last updated on 20/Aug/18 $$\mathrm{Let}\:\mathrm{P}\:\mathrm{be}\:\mathrm{an}\:\mathrm{interior}\:\mathrm{point}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{ABC}\:\mathrm{and}\:\mathrm{AP},\mathrm{BP},\mathrm{CP}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{BC}, \\ $$$$\mathrm{CA},\mathrm{AB}\:\mathrm{in}\:\mathrm{D},\mathrm{E},\mathrm{F}\:\mathrm{respectively}.\:\mathrm{Show} \\ $$$$\mathrm{that}\:\frac{\mathrm{AP}}{\mathrm{PD}}=\:\frac{\mathrm{AF}}{\mathrm{FB}}\:+\:\frac{\mathrm{AE}}{\mathrm{EC}}\:. \\ $$ Commented by rahul 19 last updated…
Question Number 42199 by rahul 19 last updated on 20/Aug/18 Answered by tanmay.chaudhury50@gmail.com last updated on 20/Aug/18 $$\left.{a}\right)\left\{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)×\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)\right\}.\left(\boldsymbol{{c}}+\boldsymbol{{a}}\right) \\ $$$$\left(\boldsymbol{{a}}×\boldsymbol{{b}}+\boldsymbol{{a}}×\boldsymbol{{c}}+\boldsymbol{{b}}×\boldsymbol{{b}}+\boldsymbol{{b}}×\boldsymbol{{c}}\right).\left(\boldsymbol{{c}}+\boldsymbol{{a}}\right) \\ $$$$\left(\boldsymbol{{a}}×\boldsymbol{{b}}\right).\boldsymbol{{c}}+\left(\boldsymbol{{b}}×\boldsymbol{{c}}\right).\boldsymbol{{a}}=\mathrm{2}{v} \\ $$$$\left[{abc}\right]=\left[{bca}\right]=\left[{cab}\right]={v} \\…
Question Number 42200 by rahul 19 last updated on 20/Aug/18 Answered by tanmay.chaudhury50@gmail.com last updated on 20/Aug/18 Commented by tanmay.chaudhury50@gmail.com last updated on 20/Aug/18 Commented…
Question Number 42180 by rahul 19 last updated on 19/Aug/18 $$\mathrm{The}\:\mathrm{median}\:\mathrm{AD}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{is}\: \\ $$$$\mathrm{bisected}\:\mathrm{at}\:\mathrm{E}\:\mathrm{and}\:\mathrm{BE}\:\mathrm{meets}\:\mathrm{AC}\:\mathrm{at}\:\mathrm{F}. \\ $$$$\mathrm{Find}\:\mathrm{AF}:\mathrm{FC}\:. \\ $$ Answered by MJS last updated on 19/Aug/18 $$\mathrm{you}\:\mathrm{can}\:\mathrm{put}\:\mathrm{any}\:\mathrm{triangle}\:{abc}\:\mathrm{in}\:\mathrm{this}\:\mathrm{position}:…
Question Number 173069 by mnjuly1970 last updated on 06/Jul/22 $$ \\ $$$$ \\ $$$$\:\:\:\Theta\:=\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} {xy}\:{e}^{\:−\left({x}+{y}\right)} {cos}\left({x}+{y}\:\right){dxdy}=\frac{\mathrm{1}}{\sigma} \\ $$$$\:\:\:\:\:\:\:\:{find}\:{the}\:\:{value}\:{of}\:\:''\:\sigma\:\:''. \\ $$$$ \\ $$…
Question Number 41606 by psyche-ace last updated on 10/Aug/18 $$\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{24}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{9}\boldsymbol{\mathrm{x}}−\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{any}}\:\boldsymbol{\mathrm{method}}.\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{real}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{x}}\:\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{satisfy}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{polynomial}} \\ $$ Answered by MJS last updated on 10/Aug/18 $${f}\left({x}\right)={x}^{\mathrm{4}}…
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Question Number 105904 by bemath last updated on 01/Aug/20 $$\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{5}−\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\:? \\ $$ Answered by bemath last updated on 01/Aug/20 Commented by Her_Majesty last updated on…
Question Number 171033 by cortano1 last updated on 06/Jun/22 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}}\:\left[\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}−\mathrm{1}}\:}−\mathrm{1}\:\right]=? \\ $$ Commented by greougoury555 last updated on 07/Jun/22 $$\:\sqrt[{\mathrm{3}}]{\frac{\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}}\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right)}{\mathrm{1}+\sqrt{\mathrm{1}−{x}}}.\frac{\sqrt{\mathrm{1}+{x}}+\mathrm{1}}{\left(\sqrt{\mathrm{1}+{x}}−\mathrm{1}\right)\left(\sqrt{\mathrm{1}+{x}}+\mathrm{1}\right)}} \\ $$$$=\:\sqrt[{\mathrm{3}}]{\frac{{x}\left(\sqrt{\mathrm{1}+{x}}+\mathrm{1}\right)}{{x}\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right)}}\:=\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{1}+{x}}+\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}−{x}}}\:} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}}…