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}} \\ $$$$\mathrm{8}\:{cm}\:{means}\:{then}\:\mathrm{8}×\mathrm{7}=\mathrm{56}\:{cm}^{\mathrm{2}} \\ $$$${green}\:{area}=\mathrm{56}−\mathrm{25}=\mathrm{31}\:{cm}^{\mathrm{2}} \\ $$…
Question Number 96785 by M±th+et+s last updated on 04/Jun/20 Answered by mr W last updated on 04/Jun/20 Commented by mr W last updated on 04/Jun/20…
Question Number 96441 by M±th+et+s last updated on 01/Jun/20 Commented by M±th+et+s last updated on 01/Jun/20 $${let}\:{p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} ….{p}_{\mathrm{60}} \:{be}\:\mathrm{60}\:{points}\:{on}\:{BC} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{60}} {\sum}}\left({AP}_{{i}}…
Question Number 96032 by i jagooll last updated on 29/May/20 Commented by i jagooll last updated on 29/May/20 $$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{an}\:\mathrm{area}\:\mathrm{240}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{DP}\:=\:\mathrm{3PC}\:,\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\: \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{region}\:? \\ $$…
Question Number 95999 by i jagooll last updated on 29/May/20 Answered by mr W last updated on 29/May/20 Commented by mr W last updated on…
Question Number 95901 by bobhans last updated on 28/May/20 $$\frac{\mathrm{1}}{\mathrm{998}!}\:+\:\frac{\mathrm{1}}{\mathrm{999}!}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{100}!}\: \\ $$ Answered by i jagooll last updated on 28/May/20 $$\frac{\mathrm{999}}{\mathrm{999}×\mathrm{998}!}\:+\:\frac{\mathrm{1}}{\mathrm{999}!}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{100}!} \\ $$$$\frac{\mathrm{1000}}{\mathrm{999}!}\:=\:\frac{\mathrm{x}^{\mathrm{3}}…
Question Number 30299 by ajfour last updated on 19/Feb/18 Commented by ajfour last updated on 19/Feb/18 $${Find}\:{maximum}\:{area}\:{of}\: \\ $$$${quadrilateral}\:{ABCD}\:{given}\:\boldsymbol{{a}},\:\boldsymbol{{r}}\:. \\ $$ Commented by ajfour last…
Question Number 95780 by john santu last updated on 27/May/20 Terms of Service Privacy Policy Contact: info@tinkutara.com