Question Number 81648 by Zainal Arifin last updated on 14/Feb/20 $$\mathrm{A}\:\mathrm{team}\:\mathrm{of}\:\mathrm{8}\:\mathrm{couples},\:\left(\mathrm{husband}\:\mathrm{and}\:\mathrm{wife}\right) \\ $$$$\mathrm{attend}\:\mathrm{a}\:\mathrm{lucky}\:\mathrm{draw}\:\mathrm{in}\:\mathrm{which}\:\mathrm{4}\:\mathrm{persons} \\ $$$$\mathrm{picked}\:\mathrm{up}\:\mathrm{for}\:\mathrm{a}\:\mathrm{prize}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{there}\:\mathrm{is}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{couple}\:\mathrm{is} \\ $$ Commented by mr W last updated…
Question Number 16107 by hamil last updated on 18/Jun/17 $$\int\:\frac{{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{3}\right)\sqrt{{x}+\mathrm{1}}}\:{dx}\:= \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 81629 by zainal tanjung last updated on 14/Feb/20 $$\mathrm{Given}\:\mathrm{vectors}\:\boldsymbol{\mathrm{x}}=\mathrm{3}\boldsymbol{\mathrm{i}}−\mathrm{6}\boldsymbol{\mathrm{j}}−\boldsymbol{\mathrm{k}},\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{i}}+\mathrm{4}\boldsymbol{\mathrm{j}}−\mathrm{3}\boldsymbol{\mathrm{k}} \\ $$$$\mathrm{and}\:\boldsymbol{\mathrm{z}}=\mathrm{3}\boldsymbol{\mathrm{i}}−\mathrm{4}\boldsymbol{\mathrm{j}}+\mathrm{12}\boldsymbol{\mathrm{k}},\:\mathrm{then}\:\mathrm{the}\:\mathrm{projection} \\ $$$$\mathrm{of}\:\boldsymbol{\mathrm{X}}×\boldsymbol{\mathrm{Y}}\:\mathrm{on}\:\mathrm{vector}\:\boldsymbol{\mathrm{Z}}\:\mathrm{is} \\ $$ Answered by mr W last updated on 14/Feb/20…
Question Number 16088 by hamil last updated on 17/Jun/17 $$\:\underset{\:\mathrm{0}} {\overset{\mathrm{1000}} {\int}}{e}^{{x}−\left[{x}\right]} {dx}\:= \\ $$ Commented by prakash jain last updated on 17/Jun/17 $${x}=\left[{x}\right]+\left\{{x}\right\}\:\mathrm{where}\:\mathrm{0}\leqslant\left\{{x}\right\}<\mathrm{1} \\…
Question Number 81615 by Zainal Arifin last updated on 14/Feb/20 $$\:\underset{\:\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:{x}\:\sqrt{\mathrm{1}+{x}}\:{dx}\:= \\ $$ Answered by TANMAY PANACEA last updated on 14/Feb/20 $${t}^{\mathrm{2}} =\mathrm{1}+{x}\rightarrow\mathrm{2}{tdt}={dx}…
Question Number 15541 by Devdeep.math.in last updated on 11/Jun/17 $$\mathrm{If}\:\theta\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{quadrant}\:\mathrm{which}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{not}\:\mathrm{true}? \\ $$ Commented by prakash jain last updated on 11/Jun/17 $$\mathrm{please}\:\mathrm{add}\:\mathrm{options}. \\ $$…
Question Number 80777 by zainal tanjung last updated on 14/Feb/20 $$\mathrm{If}\:\:^{{n}+\mathrm{2}} {C}_{\mathrm{8}} \::\:^{{n}−\mathrm{2}} {P}_{\mathrm{4}} =\:\mathrm{57}\::\:\mathrm{16},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:{n}\:\mathrm{is}\:…… \\ $$ Answered by Joel578 last updated on…
Question Number 80775 by zainal tanjung last updated on 06/Feb/20 $$\mathrm{There}\:\mathrm{are}\:{n}\:\mathrm{straight}\:\mathrm{lines}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}, \\ $$$$\mathrm{no}\:\mathrm{two}\:\mathrm{of}\:\mathrm{which}\:\mathrm{are}\:\mathrm{parallel},\:\mathrm{and}\:\mathrm{no} \\ $$$$\mathrm{three}\:\mathrm{pass}\:\mathrm{through}\:\mathrm{the}\:\mathrm{same}\:\mathrm{point}. \\ $$$$\mathrm{Their}\:\mathrm{points}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{are}\:\mathrm{joined}. \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{fresh}\:\mathrm{lines}\:\mathrm{thus}\: \\ $$$$\mathrm{obtained}\:\mathrm{is} \\ $$ Terms of…
Question Number 15207 by arnabpapu550@gmail.com last updated on 08/Jun/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:\sqrt{\mathrm{3}}\:\mathrm{cosec}\:\mathrm{20}°−\mathrm{sec}\:\mathrm{20}°\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\_\_\_\_. \\ $$ Answered by mrW1 last updated on 08/Jun/17 $$\frac{\sqrt{\mathrm{3}}}{\mathrm{sin}\:\mathrm{20}°}−\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}=\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{20}°−\mathrm{sin}\:\mathrm{20}°}{\mathrm{sin}\:\mathrm{20}°\mathrm{cos}\:\mathrm{20}°} \\ $$$$=\frac{\mathrm{4}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\mathrm{20}°−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}°\right)}{\mathrm{2sin}\:\mathrm{20}°\mathrm{cos}\:\mathrm{20}°} \\…
Question Number 15194 by arnabpapu550@gmail.com last updated on 08/Jun/17 $$\mathrm{The}\:\mathrm{equation}\:\begin{vmatrix}{{x}−{a}}&{{x}−{b}}&{{x}−{c}}\\{{x}−{b}}&{{x}−{c}}&{{x}−{a}}\\{{x}−{c}}&{{x}−{a}}&{{x}−{b}}\end{vmatrix}=\mathrm{0}, \\ $$$$\mathrm{where}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{different},\:\mathrm{is}\:\mathrm{satisfied}\:\mathrm{by} \\ $$ Commented by prakash jain last updated on 08/Jun/17 $$\:\begin{vmatrix}{{x}−{a}}&{{x}−{b}}&{{x}−{c}}\\{{x}−{b}}&{{x}−{c}}&{{x}−{a}}\\{{x}−{c}}&{{x}−{a}}&{{x}−{b}}\end{vmatrix}=\mathrm{0}, \\ $$$$\mathrm{R1}=\mathrm{R1}+\mathrm{R2}+\mathrm{R3}…