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 $$\frac{\mathrm{1}}{\mathrm{11}}×\frac{\partial^{\mathrm{4}} }{\partial^{\mathrm{2}} {x}\partial^{\mathrm{2}} {y}}\left[\frac{\mathrm{11}}{\mathrm{4}}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right]\left({exp}\left({ln}\left[\frac{\mathrm{24}}{\pi}\left\{\underset{{n}\rightarrow+\infty}…
Question Number 4225 by Rasheed Soomro last updated on 03/Jan/16 $$\:^{\bullet} \mathrm{A}\:\mathrm{kite}\:\mathrm{is}\:\mathrm{a}\:\mathrm{quadrilateral}\:\mathrm{having}\:\mathrm{two} \\ $$$$\mathrm{pairs}\:\mathrm{of}\:\mathrm{adjacent}\:\mathrm{sides}\:\mathrm{equal}. \\ $$$$\mathrm{Draw}\:\mathrm{a}\:\mathrm{semi}-\mathrm{circle}\:\mathrm{inside}\:\mathrm{it}\:\mathrm{touching} \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{using}\:\mathrm{Eucledian}\:\mathrm{tools}. \\ $$$$\:^{\bullet} \:\mathrm{Can}\:\mathrm{we}\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\:\mathrm{above}\:\mathrm{semi}-\mathrm{circle} \\ $$$$\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{laregest}\:\mathrm{possible}\:\mathrm{area}\:\mathrm{inside}\:\mathrm{the} \\ $$$$\mathrm{kite}?…
Question Number 4219 by Yozzii last updated on 02/Jan/16 Answered by prakash jain last updated on 02/Jan/16 $${f}\:'\left(\theta\right)=\underset{\delta\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left(\theta+\delta\theta\right)−{f}\left(\theta\right)}{\delta\theta} \\ $$$$\mathrm{If}\:\mathrm{limit}\:\mathrm{exits} \\ $$$$\Rightarrow\underset{\delta\theta\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left(\theta+\delta\theta\right)={f}\left(\theta\right)+\underset{\delta\theta\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\:'\left(\theta\right)\delta\theta…
Question Number 135188 by SLVR last updated on 11/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com