Question Number 4387 by Rasheed Soomro last updated on 17/Jan/16 Commented by Rasheed Soomro last updated on 17/Jan/16 $$\mathrm{In}\:\mathrm{the}\:\mathrm{trapezium}\:\mathrm{m}\angle\mathrm{A}=\mathrm{m}\angle\mathrm{B}=\frac{\pi}{\mathrm{2}}\:\mathrm{rad}. \\ $$$$\mathrm{m}\overline {\mathrm{AB}}=\mathrm{m}\overline {\mathrm{AD}}=\mathrm{x}\:\mathrm{units}\:\mathrm{and}\:\mathrm{m}\overline {\mathrm{BC}}=\mathrm{2x}\:\mathrm{units}. \\…
Question Number 135458 by Dwaipayan Shikari last updated on 13/Mar/21 Commented by Dwaipayan Shikari last updated on 13/Mar/21 Commented by Dwaipayan Shikari last updated on…
Question Number 4384 by Rasheed Soomro last updated on 16/Jan/16 $$\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…
Question Number 4374 by Rasheed Soomro last updated on 14/Jan/16 Commented by Rasheed Soomro last updated on 14/Jan/16 $$\mathcal{I}{n}\:{trapizium}\:\mathrm{ABCD} \\ $$$$\mathrm{A}\:{and}\:\mathrm{B}\:{are}\:{right}\:{angles} \\ $$$$\mathrm{AB}=\mathrm{AD}=\mathrm{x}\:\mathrm{units}\:{and}\:\mathrm{BC}=\mathrm{2x}\:\mathrm{units}. \\ $$$${The}\:{trapizium}\:\mathrm{ABCD}\:{has}\:{been}\:…
Question Number 4356 by Rasheed Soomro last updated on 12/Jan/16 Commented by Rasheed Soomro last updated on 12/Jan/16 $$\mathcal{D}{etermine}\:{area}\:{of}\:{above}\:{closed} \\ $$$${figure}\:\:{in}\:{the}\:{easiest}\:{way}. \\ $$ Answered by…
Question Number 135421 by benjo_mathlover last updated on 13/Mar/21 Commented by mr W last updated on 13/Mar/21 $${radius}\:{of}\:{semicircle} \\ $$$${R}={a}+{b}−\sqrt{\mathrm{2}{ab}} \\ $$ Commented by EDWIN88…
Question Number 4329 by Rasheed Soomro last updated on 10/Jan/16 Commented by prakash jain last updated on 10/Jan/16 $$\mathrm{square}/\mathrm{8}+\mathrm{traingle}/\mathrm{4}=\mathrm{12}\:\mathrm{parts} \\ $$$$\mathrm{each}\:\mathrm{part}\:\mathrm{gets}\:\mathrm{3}. \\ $$ Answered by…
Question Number 4326 by pedro pablo last updated on 10/Jan/16 $${hola} \\ $$ Answered by pedro pablo last updated on 10/Jan/16 Answered by pedro pablo…
Question Number 4318 by Rasheed Soomro last updated on 10/Jan/16 Commented by 123456 last updated on 10/Jan/16 $${x}/\mathrm{2} \\ $$ Commented by prakash jain last…
Question Number 4268 by Momeen last updated on 06/Jan/16 $$\mathrm{1}+\mathrm{2}= \\ $$ Answered by Yozzii last updated on 06/Jan/16 $$\frac{\mathrm{1}}{\mathrm{11}}×\frac{\partial^{\mathrm{4}} }{\partial^{\mathrm{2}} {x}\partial^{\mathrm{2}} {y}}\left[\frac{\mathrm{11}}{\mathrm{4}}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right]\left({exp}\left({ln}\left[\frac{\mathrm{24}}{\pi}\left\{\underset{{n}\rightarrow+\infty}…