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Question Number 64406 by rajesh4661kumar@gamil.com last updated on 17/Jul/19 Commented by Tony Lin last updated on 17/Jul/19 $${area}\:{of}\:\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}\mid\begin{vmatrix}{{x}_{\mathrm{2}} −{x}_{\mathrm{1}} \:\:{y}_{\mathrm{2}} −{y}_{\mathrm{1}} }\\{{x}_{\mathrm{3}} −{x}_{\mathrm{1}} \:\:{y}_{\mathrm{3}} −{y}_{\mathrm{1}}…
Question Number 129886 by zarawan last updated on 23/Jan/21 $${Find}\:{Null}\:{space}\:{of}\:{the}\:{following}\:{matrix}\:{and}\:{also}\:{find}\:{basis}\:{for}\:{the}\:{null}\:{space}. \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{0}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}&{\mathrm{1}}\end{bmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 64124 by ANTARES VY last updated on 13/Jul/19 $$\sqrt{\mathrm{2014}}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{4029}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{1}} <\boldsymbol{\mathrm{x}}_{\mathrm{2}} <\boldsymbol{\mathrm{x}}_{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}_{\mathrm{1}} +\boldsymbol{\mathrm{x}}_{\mathrm{3}} \right)=? \\ $$ Terms…
Question Number 64012 by aditya@345 last updated on 12/Jul/19 Answered by MJS last updated on 12/Jul/19 $$\begin{vmatrix}{\mathrm{3}}&{−\mathrm{2}}&{\mathrm{sin}\:\mathrm{3}\theta}\\{−\mathrm{7}}&{\mathrm{8}}&{\mathrm{cos}\:\mathrm{2}\theta}\\{−\mathrm{11}}&{\mathrm{14}}&{\mathrm{2}}\end{vmatrix}=\mathrm{0} \\ $$$$\mathrm{20}−\mathrm{20cos}\:\mathrm{2}\theta\:−\mathrm{10sin}\:\mathrm{3}\theta\:=\mathrm{0} \\ $$$$\mathrm{2cos}\:\mathrm{2}\theta\:+\mathrm{sin}\:\mathrm{3}\theta\:−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{1}−\mathrm{2sin}\:\theta\right)\left(\mathrm{3}+\mathrm{2sin}\:\theta\right)\mathrm{sin}\:\theta\:=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\:=−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:\theta\notin\mathbb{R}…
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Question Number 127143 by peter frank last updated on 27/Dec/20 Commented by Dwaipayan Shikari last updated on 27/Dec/20 $$\mathrm{042} \\ $$ Answered by Ar Brandon…