Question Number 108776 by bobhans last updated on 19/Aug/20 $$\:\overset{{bobhans}} {\vdots} \\ $$$$\int\:\frac{{dx}}{\:\sqrt{{x}\sqrt{{x}}\:−{x}^{\mathrm{2}} }}\:=\:? \\ $$ Answered by john santu last updated on 19/Aug/20 $$\:\:\:\:\:\frac{\multimap\mathcal{JS}\multimap}{\heartsuit}…
Question Number 42711 by rahul 19 last updated on 01/Sep/18 $$\mathrm{If}\:\mathrm{2}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{are}\:\hat {\mathrm{i}}+\mathrm{2}\hat {\mathrm{j}}\:\mathrm{and} \\ $$$$\hat {\mathrm{i}}+\hat {\mathrm{k}}\:,\:\mathrm{then}\:\mathrm{find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{third}\:\mathrm{side}\:? \\ $$ Answered by MJS last updated on…
Question Number 42528 by rahul 19 last updated on 27/Aug/18 $$\mathrm{Solve}\:: \\ $$$$\frac{\mathrm{p}+\mathrm{2}}{\mathrm{2p}+\mathrm{1}}\:=\:\frac{\mathrm{q}+\mathrm{2p}}{\mathrm{2q}+\mathrm{p}}\:=\:\frac{\mathrm{1}+\mathrm{2q}}{\mathrm{2}+\mathrm{q}}\:=\:\lambda. \\ $$$$\mathrm{Find}\:\left(\mathrm{p},\mathrm{q}\right)\:? \\ $$ Answered by math1967 last updated on 27/Aug/18 $$\lambda=\frac{{p}+\mathrm{2}+{q}+\mathrm{2}{p}+\mathrm{1}+\mathrm{2}{q}}{\mathrm{2}{p}+\mathrm{1}+\mathrm{2}{q}+{p}+\mathrm{2}+{q}}=\frac{\mathrm{3}\left({p}+{q}+\mathrm{1}\right)}{\mathrm{3}\left({p}+{q}+\mathrm{1}\right)}=\mathrm{1}…
Question Number 42521 by rahul 19 last updated on 27/Aug/18 $$\mathrm{Let}\:\overset{\rightarrow\:} {\mathrm{a}},\:\overset{\rightarrow} {\mathrm{b}}\:,\:\overset{\rightarrow} {\mathrm{c}}\:\mathrm{be}\:\mathrm{three}\:\mathrm{unit}\:\mathrm{vectors} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{3}\overset{\rightarrow} {\mathrm{a}}+\mathrm{4}\overset{\rightarrow} {\mathrm{b}}+\mathrm{5}\overset{\rightarrow} {\mathrm{c}}\:=\:\mathrm{0}.\:\mathrm{Then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\overset{\rightarrow\:} {\mathrm{a}},\:\overset{\rightarrow} {\mathrm{b}},\overset{\rightarrow} {\mathrm{c}}\:\mathrm{are}\:\mathrm{coplanar}. \\…
Question Number 42357 by preet last updated on 24/Aug/18 Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18 $$\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 $$\mathrm{you}\:\mathrm{can}\:\mathrm{put}\:\mathrm{any}\:\mathrm{triangle}\:{abc}\:\mathrm{in}\:\mathrm{this}\:\mathrm{position}:…
Question Number 173069 by mnjuly1970 last updated on 06/Jul/22 $$ \\ $$$$ \\ $$$$\:\:\:\Theta\:=\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} {xy}\:{e}^{\:−\left({x}+{y}\right)} {cos}\left({x}+{y}\:\right){dxdy}=\frac{\mathrm{1}}{\sigma} \\ $$$$\:\:\:\:\:\:\:\:{find}\:{the}\:\:{value}\:{of}\:\:''\:\sigma\:\:''. \\ $$$$ \\ $$…