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Author: Tinku Tara

I-A-5-1-B-3-5-C-5-2-ar-ABC-II-A-5-3-B-2-5-C-5-3-D-4-3-ar-ABCD-shortest-solution-

Question Number 204062 by BaliramKumar last updated on 05/Feb/24 $$\mathrm{I}.\:\:\:\:\:\:\:\mathrm{A}\left(−\mathrm{5},\:−\mathrm{1}\right);\:\mathrm{B}\left(\mathrm{3},\:−\mathrm{5}\right);\:\mathrm{C}\left(\mathrm{5},\:\mathrm{2}\right)\:\:\:\:\:\:{ar}\left(\bigtriangleup\mathrm{ABC}\right)\:=\:? \\ $$$$\mathrm{II}.\:\:\:\:\:\mathrm{A}\left(\mathrm{5},\:\mathrm{3}\right);\:\mathrm{B}\left(\mathrm{2},\:\mathrm{5}\right);\:\mathrm{C}\left(−\mathrm{5},\:\mathrm{3}\right);\:\mathrm{D}\left(−\mathrm{4},\:−\mathrm{3}\right)\:\:\:\:\:\:\:{ar}\left(\Box\mathrm{ABCD}\right)\:=\:? \\ $$$$\mathrm{shortest}\:\mathrm{solution}\: \\ $$ Answered by mr W last updated on 05/Feb/24 $${I}.…

x-2y-z-8-3x-2y-z-10-4x-3y-2z-4-Solve-with-the-help-of-matrix-

Question Number 204056 by hardmath last updated on 05/Feb/24 $$\begin{cases}{\mathrm{x}\:+\:\mathrm{2y}\:+\:\mathrm{z}\:=\:\mathrm{8}}\\{\mathrm{3x}\:+\:\mathrm{2y}\:+\:\mathrm{z}\:=\:\mathrm{10}}\\{\mathrm{4x}\:+\:\mathrm{3y}\:−\:\mathrm{2z}\:=\:\mathrm{4}}\end{cases} \\ $$$$\mathrm{Solve}\:\mathrm{with}\:\mathrm{the}\:\mathrm{help}\:\mathrm{of}\:\mathrm{matrix} \\ $$ Answered by AST last updated on 05/Feb/24 $$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{4}}&{\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}\\{\mathrm{10}}\\{\mathrm{4}}\end{bmatrix}\underset{{R}_{\mathrm{3}} −\mathrm{4}{R}_{\mathrm{1}} } {\overset{{R}_{\mathrm{2}}…

y-sinx-cos-3-x-find-y-

Question Number 204054 by hardmath last updated on 05/Feb/24 $$\mathrm{y}\:=\:\sqrt{\mathrm{sinx}}\:+\:\mathrm{cos}^{\mathrm{3}} \mathrm{x} \\ $$$$\mathrm{find}:\:\:\mathrm{y}^{'} \:=\:? \\ $$ Answered by AST last updated on 05/Feb/24 $${y}'=\frac{{cos}\left({x}\right)}{\:\mathrm{2}\sqrt{{sinx}}}−\mathrm{3}{sin}\left({x}\right){cos}^{\mathrm{2}} {x}…

y-2-3-arctg-x-4-find-y-

Question Number 204055 by hardmath last updated on 05/Feb/24 $$\mathrm{y}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{arctg}\:\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\mathrm{find}:\:\:\mathrm{y}^{'} \:=\:? \\ $$ Answered by AST last updated on 05/Feb/24 $${y}^{'} =\frac{\mathrm{2}}{\mathrm{3}}\left[\frac{{d}}{{dx}}{tan}^{−\mathrm{1}}…

sin-2-2-sin-2-90-gt-gt-0-

Question Number 204072 by Red1ight last updated on 05/Feb/24 $$\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}}=\mathrm{sin}\:\mathrm{2}\theta?\:\left(\mathrm{90}°>\theta>\mathrm{0}°\right) \\ $$ Answered by AST last updated on 05/Feb/24 $${sin}^{\mathrm{2}} \theta=\mathrm{4}{sin}\theta{cos}\theta\Rightarrow{tan}\theta=\mathrm{4}\Rightarrow\theta={tan}^{−\mathrm{1}} \left(\mathrm{4}\right)\approx\mathrm{75}.\mathrm{96}° \\ $$…