Question Number 204101 by necx122 last updated on 06/Feb/24 $${Find}\:{the}\:{possible}\:{values}\:{for}\:{x}\:{and}\:{y}\:{if} \\ $$$$\mathrm{10}{cosx}\:+\:\mathrm{12}{cos}\left({x}+{y}\right)=\mathrm{5} \\ $$$$\mathrm{10}{sinx}\:+\:\mathrm{12}{sin}\left({x}+{y}\right)=\mathrm{20}.\mathrm{66} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 204103 by MMMMMLyv last updated on 06/Feb/24 $$ \\ $$$$ \\ $$$$ \\ $$विद्युत आवेश का SI मात्रक कुलम (C) होता है $$…
Question Number 204129 by hardmath last updated on 06/Feb/24 $$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{x}\:,\:\mathrm{y}\:\in\:\mathbb{R} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:=\:\mathrm{23} \\ $$$$\mathrm{ax}\:+\:\mathrm{by}\:=\:\mathrm{79} \\ $$$$\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{by}^{\mathrm{2}} \:=\:\mathrm{217} \\ $$$$\mathrm{ax}^{\mathrm{3}} \:+\:\mathrm{by}^{\mathrm{3}} \:=\:\mathrm{661} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{by}^{\mathrm{4}}…
Question Number 204062 by BaliramKumar last updated on 05/Feb/24 $$\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 $${I}.…
Question Number 204056 by hardmath last updated on 05/Feb/24 $$\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 $$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{4}}&{\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}\\{\mathrm{10}}\\{\mathrm{4}}\end{bmatrix}\underset{{R}_{\mathrm{3}} −\mathrm{4}{R}_{\mathrm{1}} } {\overset{{R}_{\mathrm{2}}…
Question Number 204054 by hardmath last updated on 05/Feb/24 $$\mathrm{y}\:=\:\sqrt{\mathrm{sinx}}\:+\:\mathrm{cos}^{\mathrm{3}} \mathrm{x} \\ $$$$\mathrm{find}:\:\:\mathrm{y}^{'} \:=\:? \\ $$ Answered by AST last updated on 05/Feb/24 $${y}'=\frac{{cos}\left({x}\right)}{\:\mathrm{2}\sqrt{{sinx}}}−\mathrm{3}{sin}\left({x}\right){cos}^{\mathrm{2}} {x}…
Question Number 204055 by hardmath last updated on 05/Feb/24 $$\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 $${y}^{'} =\frac{\mathrm{2}}{\mathrm{3}}\left[\frac{{d}}{{dx}}{tan}^{−\mathrm{1}}…
Question Number 204081 by esmaeil last updated on 05/Feb/24 $${what}\:{is}\:{the}\:{area}\:{of}\:{the}\:{largest}\:{square} \\ $$$${that}\:{can}\:{be}\:{enclosed}\:{in}\:{a}\:{triangle} \\ $$$${with}\:{an}\:{area}\:{of}\:\mathrm{1}? \\ $$ Answered by mr W last updated on 06/Feb/24 Commented…
Question Number 204082 by depressiveshrek last updated on 05/Feb/24 $$\sqrt{\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }+{x}}{{x}}}−\sqrt{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }−{x}^{\mathrm{2}} }=\mathrm{5} \\ $$ Commented by Frix last updated on 05/Feb/24 $${x}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{119}}}…