Question Number 97888 by M±th+et+s last updated on 10/Jun/20 $${how}\:{to}\:{prove}\:\left(\left({the}\:{volumn}\:{of}\right.\right. \\ $$$$\left.{d}\left.{imensional}\:{sphare}\right)\right)\:{formula} \\ $$$${V}_{{N}} \left({R}\right)=\frac{\pi^{{N}/\mathrm{2}} }{\Gamma\left(\frac{{N}}{\mathrm{2}}+\mathrm{1}\right)}\:{R}^{{N}} \\ $$$$ \\ $$ Commented by EmericGent last updated…
Question Number 97858 by bemath last updated on 10/Jun/20 Commented by bemath last updated on 10/Jun/20 $$\mathrm{find}\:\mathrm{shaded}\:\mathrm{area}\:\mathrm{region} \\ $$ Answered by bobhans last updated on…
Question Number 97497 by bemath last updated on 08/Jun/20 Answered by smridha last updated on 08/Jun/20 $$\mathrm{4}\left(\boldsymbol{\pi}−\sqrt{\mathrm{3}}\right)\left(\boldsymbol{{unit}}\right)^{\mathrm{2}} \\ $$ Commented by john santu last updated…
Question Number 97494 by bemath last updated on 08/Jun/20 Answered by smridha last updated on 08/Jun/20 $$\mathrm{4}^{\mathrm{2}} =\mathrm{16}{simple} \\ $$ Commented by bemath last updated…
Question Number 97088 by bemath last updated on 06/Jun/20 Answered by Fikret last updated on 06/Jun/20 Commented by mr W last updated on 06/Jun/20 $${maybe}\:{you}\:{are}\:{right}\:{sir}…
Question Number 97068 by bemath last updated on 06/Jun/20 Commented by bemath last updated on 06/Jun/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{dimension}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle} \\ $$$$\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{so}\:\mathrm{area}\:\mathrm{rectangle} \\ $$$$+\:\mathrm{area}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{maximum} \\ $$ Answered by…
Question Number 96922 by M±th+et+s last updated on 05/Jun/20 Commented by M±th+et+s last updated on 05/Jun/20 $${prove}\:{that}:\:{r}_{\mathrm{1}} +{r}_{\mathrm{3}} +{r}_{\mathrm{5}} ={r}_{\mathrm{2}} +{r}_{\mathrm{4}} +_{\mathrm{6}} \\ $$ Commented…
Question Number 96870 by john santu last updated on 05/Jun/20 Answered by Sourav mridha last updated on 05/Jun/20 $$\boldsymbol{{A}}=\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{\pi}\left(\mathrm{2}\right)^{\mathrm{2}} −\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{4}−\boldsymbol{{x}}^{\mathrm{2}} }\:\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:=\boldsymbol{\pi}−\left[\frac{\boldsymbol{{x}}\sqrt{\mathrm{4}−\boldsymbol{{x}}^{\mathrm{2}}…
Question Number 96866 by bobhans last updated on 05/Jun/20 Commented by john santu last updated on 05/Jun/20 $$\mathrm{ratio}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{5}}×\mathrm{4}\sqrt{\mathrm{5}}}{\left(\mathrm{4}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\left(\mathrm{8}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\left(\mathrm{4}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\left(\mathrm{4}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}^{\mathrm{2}} \right)}\:=\:\mathrm{1}:\:\mathrm{5}…
Question Number 96823 by john santu last updated on 05/Jun/20 Commented by mr W last updated on 05/Jun/20 $$\mathrm{1}\:{cm}\:{means}\:\mathrm{32}−\mathrm{25}=\mathrm{7}\:{cm}^{\mathrm{2}} \\ $$$$\mathrm{8}\:{cm}\:{means}\:{then}\:\mathrm{8}×\mathrm{7}=\mathrm{56}\:{cm}^{\mathrm{2}} \\ $$$${green}\:{area}=\mathrm{56}−\mathrm{25}=\mathrm{31}\:{cm}^{\mathrm{2}} \\ $$…