Question Number 212460 by 281981 last updated on 14/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212457 by 281981 last updated on 14/Oct/24 Answered by som(math1967) last updated on 14/Oct/24 $$\:\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{b}}&{{c}}&{{a}}\\{{c}}&{{a}}&{{b}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow{a}\left({bc}−{a}^{\mathrm{2}} \right)−{b}\left({b}^{\mathrm{2}} −{ca}\right)+{c}\left({ab}−{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}}…
Question Number 212445 by ChantalYah last updated on 13/Oct/24 Commented by Frix last updated on 13/Oct/24 $$\mathrm{80cos}\:{A}\:?\:\mathrm{150sin}\:{A}\:=\mathrm{13} \\ $$$$ \\ $$$$\mathrm{80cos}\:{A}\:−\mathrm{150sin}\:{A}\:=\mathrm{13} \\ $$$$−\mathrm{170sin}\:\left({A}−\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\right)\:=\mathrm{13} \\…
Question Number 212384 by a.lgnaoui last updated on 14/Oct/24 $$\mathrm{Reponse}\:\mathrm{a}\:\mathrm{la}\:\mathrm{question}\:\mathrm{N}° \\ $$$$\mathrm{Q212291}\:\:\:\:\:\:\mathrm{S}\left(\mathrm{ABCD}\right)=\:\:\mathrm{66},\mathrm{69} \\ $$ Commented by hardmath last updated on 12/Oct/24 $$\mathrm{dear}\:\mathrm{professor},\:\mathrm{answer}:\:\mathrm{90} \\ $$ Terms…
Question Number 212200 by wajed last updated on 06/Oct/24 $$\angle\:{T}\:\:=\:\mathrm{65}°\:\:\:\: \\ $$$$\angle\:{P}\:\:=\:\mathrm{95}°\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212011 by a.lgnaoui last updated on 28/Sep/24 $$\boldsymbol{\mathrm{D}}\mathrm{eterminer}:\:\:\boldsymbol{\mathrm{R}}\mathrm{1}\:\:\:\boldsymbol{\mathrm{R}}\mathrm{2}\:\:\:\boldsymbol{\mathrm{R}}\mathrm{3} \\ $$$$\boldsymbol{\mathrm{pour}}\:\:\:\boldsymbol{\mathrm{b}}=\mathrm{12}\boldsymbol{\mathrm{cm}}\:\:\:\: \\ $$$$\boldsymbol{\mathrm{EF}}\://\:\boldsymbol{\mathrm{MN}}\:;\:\:\boldsymbol{\mathrm{EF}}\:\boldsymbol{\mathrm{Tangent}}\:\boldsymbol{\mathrm{aux}}\:\boldsymbol{\mathrm{cercles}}:\: \\ $$$$\:\:\boldsymbol{\mathrm{C}}\mathrm{1}\left(\boldsymbol{\mathrm{R}}\mathrm{1}\right)\:\:\:\boldsymbol{\mathrm{C}}\mathrm{2}\left(\boldsymbol{\mathrm{R}}\mathrm{2}\right)\:\:;\:\:\:\boldsymbol{\mathrm{EF}}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\boldsymbol{\mathrm{MN}}=\boldsymbol{\mathrm{b}} \\ $$$$\boldsymbol{\mathrm{MN}}:\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{au}}\:\boldsymbol{\mathrm{cercle}}\:\boldsymbol{\mathrm{C}}\mathrm{2} \\ $$$$\boldsymbol{\mathrm{OM}}=\boldsymbol{\mathrm{ON}}=\frac{\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\measuredangle\mathrm{MON}=\mathrm{2}\boldsymbol{\mathrm{x}}\:\:\: \\ $$$$ \\ $$ Commented…
Question Number 211961 by Nadirhashim last updated on 25/Sep/24 $$\:\:\:\boldsymbol{{if}}\:\:\:\mathrm{7}^{\boldsymbol{{sin}}^{\mathrm{2}\:} \boldsymbol{{x}}} +\:\mathrm{7}^{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}} =\:\mathrm{8}\:\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$ Answered by efronzo1 last updated on 25/Sep/24 $$\:\:\mathrm{7}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}…
Question Number 211956 by Durganand last updated on 25/Sep/24 Answered by Frix last updated on 25/Sep/24 $$\mathrm{tan}\:\alpha\:={t}\:\:\:\:\:\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\:\:\:\:\mathrm{sin}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{2}−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha} \\…
Question Number 211732 by BaliramKumar last updated on 19/Sep/24 Answered by TonyCWX08 last updated on 19/Sep/24 $${Using}\:{quadratic}\:{formula} \\ $$$${x}=\frac{{sin}^{\mathrm{2}} \theta\pm\sqrt{{sin}^{\mathrm{4}} \theta\:+\:\mathrm{4}{cos}^{\mathrm{2}} \theta}}{\mathrm{2}} \\ $$$$ \\…
Question Number 211672 by MATHEMATICSAM last updated on 15/Sep/24 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:{c}\:\mathrm{is} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{side}\:{b}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{2tanC}\:+\:\mathrm{tanA}\:=\:\mathrm{0}. \\ $$ Commented by Frix last updated on 16/Sep/24 $$\mathrm{If}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{a}\:{side}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to} \\…