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Category: Mensuration

Question-152985

Question Number 152985 by ajfour last updated on 03/Sep/21 Commented by mr W last updated on 03/Sep/21 $${area}\:{in}\:{first}\:{quadrant}\: \\ $$$$+\:{area}\:{in}\:{second}\:{quadrant}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{of}\:{rectangle}\:=\:{constant} \\ $$$${i}.{e}.\:{when}\:{area}\:{in}\:{first}\:{quadrant}\:{is} \\…

Question-152829

Question Number 152829 by liberty last updated on 01/Sep/21 Answered by MJS_new last updated on 02/Sep/21 $${z}^{\mathrm{2}} =\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\Rightarrow\:{f}\left({x},\:{y},\:{z}\right)={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}+\mathrm{2}{y}−\mathrm{1} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{obviously}\:{x}\geqslant\mathrm{0}\wedge{y}\geqslant\mathrm{0}\:\mathrm{to}\:\mathrm{get}\:\mathrm{a}\:\mathrm{maximum}…

find-the-area-of-the-region-enclosed-by-the-polar-curve-r-4-2-cos-

Question Number 87146 by john santu last updated on 03/Apr/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\: \\ $$$$\mathrm{enclosed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{curve}\: \\ $$$$\mathrm{r}\:=\:\mathrm{4}\:+\:\mathrm{2}\:\mathrm{cos}\:\theta\:? \\ $$ Answered by jagoll last updated on 03/Apr/20 Commented…

Question-87133

Question Number 87133 by jagoll last updated on 03/Apr/20 Commented by jagoll last updated on 03/Apr/20 $$\mathrm{dear}\:\mathrm{miss}\:\mathrm{tawa}. \\ $$$$\angle\:\mathrm{MEA}\:=\:\mathrm{45}^{\mathrm{o}} \:,\:\angle\mathrm{CAB}\:=\:\mathrm{19}^{\mathrm{o}} \\ $$$$\mathrm{find}\:\angle\mathrm{CPB}\:=\:? \\ $$ Commented…

Question-87027

Question Number 87027 by john santu last updated on 02/Apr/20 Answered by som(math1967) last updated on 05/Apr/20 $$\angle{DCO}={alt}\angle{OAP}=\theta\left({let}\right) \\ $$$$\angle{COQ}=\angle{OAP}=\theta \\ $$$${again}\:\angle{ADO}=\angle{DCO}=\theta{ns} \\ $$$$\left[\bigtriangleup{ADO}\sim\bigtriangleup{DCO}\right] \\…

Question-151934

Question Number 151934 by john_santu last updated on 24/Aug/21 Answered by MJS_new last updated on 25/Aug/21 $$\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\left(\mathrm{3}+\mathrm{4}\sqrt{{x}+\mathrm{1}}\right){x} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{11}}{\mathrm{18}}+\frac{\sqrt{\mathrm{13}}}{\mathrm{18}} \\ $$$${x}_{\mathrm{2}}…