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Author: Tinku Tara

A-circle-of-radius-r-1-has-been-divided-into-two-parts-of-equal-area-by-an-arc-having-center-on-the-circle-Determine-the-radius-r-2-of-the-arc-

Question Number 4384 by Rasheed Soomro last updated on 16/Jan/16 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{r}_{\mathrm{1}} \:\mathrm{has}\:\mathrm{been}\:\mathrm{divided} \\ $$$$\mathrm{into}\:\mathrm{two}\:\mathrm{parts}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{area}, \\ $$$$\mathrm{by}\:\mathrm{an}\:\mathrm{arc}\:\mathrm{having}\:\mathrm{center}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{radius}\left(\mathrm{r}_{\mathrm{2}} \right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{arc}. \\ $$ Commented by Rasheed Soomro…

xy-3-y-Find-dy-dx-

Question Number 69912 by naka3546 last updated on 29/Sep/19 $$\sqrt{{xy}}\:\:+\:\:\mathrm{3}\:\:=\:\:{y} \\ $$$${Find}\:\:\frac{{dy}}{{dx}}\:\:. \\ $$ Answered by MJS last updated on 29/Sep/19 $$\frac{{y}}{\mathrm{2}\sqrt{{xy}}}{dx}+\frac{{x}}{\mathrm{2}\sqrt{{xy}}}{dy}={dy} \\ $$$$\frac{{y}}{\mathrm{2}\sqrt{{xy}}}{dx}=\left(\mathrm{1}−\frac{{x}}{\mathrm{2}\sqrt{{xy}}}\right){dy} \\…

Question-4374

Question Number 4374 by Rasheed Soomro last updated on 14/Jan/16 Commented by Rasheed Soomro last updated on 14/Jan/16 $$\mathcal{I}{n}\:{trapizium}\:\mathrm{ABCD} \\ $$$$\mathrm{A}\:{and}\:\mathrm{B}\:{are}\:{right}\:{angles} \\ $$$$\mathrm{AB}=\mathrm{AD}=\mathrm{x}\:\mathrm{units}\:{and}\:\mathrm{BC}=\mathrm{2x}\:\mathrm{units}. \\ $$$${The}\:{trapizium}\:\mathrm{ABCD}\:{has}\:{been}\:…

Notifications-are-working-now-We-are-still-working-on-other-issues-No-app-update-is-needed-The-changes-were-only-on-server-side-

Question Number 69906 by Tinku Tara last updated on 28/Sep/19 $$\mathrm{Notifications}\:\mathrm{are}\:\mathrm{working}\:\mathrm{now}. \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{still}\:\mathrm{working}\:\mathrm{on}\:\mathrm{other}\:\mathrm{issues}. \\ $$$$ \\ $$$$\mathrm{No}\:\mathrm{app}\:\mathrm{update}\:\mathrm{is}\:\mathrm{needed}.\:\mathrm{The}\:\mathrm{changes} \\ $$$$\mathrm{were}\:\mathrm{only}\:\mathrm{on}\:\mathrm{server}\:\mathrm{side}. \\ $$ Commented by ajfour last…