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} \\ $$$$\mathrm{If}\:\mathrm{limit}\:\mathrm{exits} \\ $$$$\Rightarrow\underset{\delta\theta\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left(\theta+\delta\theta\right)={f}\left(\theta\right)+\underset{\delta\theta\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\:'\left(\theta\right)\delta\theta…
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Question Number 135176 by SLVR last updated on 11/Mar/21 $${The}\:{chord}\:{of}\:{contact}\:{of}\:{tangents}\:{fromP} \\ $$$${to}\:{a}\:{cicle}\:{pass}\:{through}\:{Q}.{If}\:{lengths}\:{of}\:{tangents}\:{from}\:{P},{Q} \\ $$$${are}\:{l}_{\mathrm{1}} ,{l}_{\mathrm{2}} \:{then}\:{PQ}\:{is}\:\sqrt{{l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} }\:{how}…{kindly}\:{tell} \\ $$ Terms of Service…
Question Number 69643 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by MJS last updated on 26/Sep/19 $${l}+{w}=\mathrm{8} \\ $$$${l}^{\mathrm{2}} +{w}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} \\ $$$$\Rightarrow\:{l}=\mathrm{4}\pm\sqrt{\mathrm{2}}\wedge{w}=\mathrm{4}\mp\sqrt{\mathrm{2}} \\ $$$$\mathrm{perimeter}\:{P}=\mathrm{10}+\mathrm{3}\pi…
Question Number 4080 by Rasheed Soomro last updated on 27/Dec/15 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{inside}\:\mathrm{a}\:\mathrm{square},\:\mathrm{a}\:\mathrm{semi}-\mathrm{circle}, \\ $$$$\mathrm{which}\:\mathrm{touches}\:\mathrm{all}\:\mathrm{the}\:\mathrm{four}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{square},\:\mathrm{is}\:\mathrm{possible}\:\mathrm{with}\:\boldsymbol{\mathrm{ruler}}\:\mathrm{and}\:\boldsymbol{\mathrm{compass}}. \\ $$ Commented by Rasheed Soomro last updated on 29/Dec/15…
Question Number 4077 by Rasheed Soomro last updated on 28/Dec/15 $$\mathrm{For}\:\mathrm{two}\:\boldsymbol{\mathrm{co}}-\boldsymbol{\mathrm{planer}}\:\mathrm{circles}\:\mathrm{to}\:\mathrm{be}\:\mathrm{tangent} \\ $$$$\:\mathrm{necessary}\:\mathrm{and}\:\mathrm{sufficient}\:\mathrm{condition}\:\mathrm{is}, \\ $$$$\mathrm{I}\:\mathrm{think} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\: \\ $$$$\:\:\:\:\mathrm{circles}\:\:\mathrm{must}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to}\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\mathrm{or}\:\mid\boldsymbol{\mathrm{r}}_{\mathrm{1}} −\boldsymbol{\mathrm{r}}_{\mathrm{2}} \mid\:, \\ $$$$\:\:\:\:\:\mathrm{where}\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}}…
Question Number 69606 by ABCDEFGHljhuvhgh last updated on 25/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 69609 by TawaTawa last updated on 25/Sep/19 Commented by Prithwish sen last updated on 25/Sep/19 $$\mathrm{I}\:\mathrm{think} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{inner}\:\mathrm{section}\:=\:\pi\left(\mathrm{75}\right)^{\mathrm{2}} \mathrm{sq}.\:\mathrm{cm} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{section}\:=\:\pi\left(\mathrm{100}−\mathrm{75}\right)\left(\mathrm{100}+\mathrm{75}\right) \\ $$$$=\:\pi×\mathrm{25}×\mathrm{175}\:\mathrm{sq}.\:\mathrm{cm}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…