Question Number 47515 by ajfour last updated on 11/Nov/18 Commented by ajfour last updated on 11/Nov/18 $${Find}\:\alpha\:{in}\:{terms}\:{of}\:{R}\:{if}\:{both} \\ $$$${coloured}\:{areas}\:{are}\:{equal}. \\ $$ Answered by mr W…
Question Number 178339 by a.lgnaoui last updated on 15/Oct/22 $${Determiner}\:{le}\:{temps}\:{necessaire}\:{pour} \\ $$$${remplir}\:{le}\:{verre}\:{a}\:{la}\:{hauteur}\:{h}. \\ $$ Commented by a.lgnaoui last updated on 15/Oct/22 Commented by CElcedricjunior last…
Question Number 112780 by ajfour last updated on 09/Sep/20 Commented by ajfour last updated on 09/Sep/20 $${If}\:\:{regions}\:{A},\:{B},\:{C},\:{D}\:{have}\:{equal} \\ $$$${areas},\:{find}\:{P}\:\left({h},{k}\right)\:{in}\:{terms}\:{of}\:{the} \\ $$$${square}\:{side}\:\boldsymbol{{a}}. \\ $$ Answered by…
Question Number 112686 by ajfour last updated on 09/Sep/20 Commented by ajfour last updated on 14/Sep/20 $${If}\:\:{the}\:{material}\:{density}\:{of}\:{the}\:{solid} \\ $$$${sphere}\:{varies}\:{as}\:\:\boldsymbol{\rho}\:=\:\rho_{\mathrm{0}} \left(\frac{{r}}{{R}}\right)\mathrm{sin}\:\phi\mathrm{cos}\:^{\mathrm{2}} \theta\:, \\ $$$${find}\:{the}\:{average}\:{density}\:{in}\:{terms} \\ $$$${of}\:\rho_{\mathrm{0}}…
Question Number 47145 by ajfour last updated on 05/Nov/18 Commented by ajfour last updated on 05/Nov/18 $${Q}.\mathrm{47113}\:\:\:{Find}\:{volume}\:{of} \\ $$$${pyramid}\:{in}\:{terms}\:{of}\:{a},{b},{c},{p},{q},{r}. \\ $$ Answered by ajfour last…
Question Number 47010 by ajfour last updated on 03/Nov/18 Commented by ajfour last updated on 03/Nov/18 $${Find}\:{maximum}\:{rectangular} \\ $$$${area}\:{and}\:\boldsymbol{{l}}\:\&\:\boldsymbol{{h}}\:\left({then}\right),\:{in}\:{terms} \\ $$$${of}\:{sides}\:{a},{b},{c}\:\:{of}\:\bigtriangleup{ABC}. \\ $$$$\left({with}\:{QR}\:\:{along}\:{BC}\right) \\ $$…
Question Number 112467 by Aina Samuel Temidayo last updated on 08/Sep/20 Answered by 1549442205PVT last updated on 08/Sep/20 $$\mathrm{Side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{Width}\:\mathrm{of}\:\mathrm{rectangle}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{2a}}{\:\sqrt{\mathrm{2}}}=\mathrm{a}\sqrt{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{rectangle}\:\mathrm{is}\:\mathrm{S}_{\mathrm{1}} =\mathrm{2a}.\mathrm{a}\sqrt{\mathrm{2}}=\mathrm{2a}^{\mathrm{2}} \sqrt{\mathrm{2}}\:\left(\mathrm{1}\right)…
Question Number 177578 by BaliramKumar last updated on 07/Oct/22 Commented by BaliramKumar last updated on 07/Oct/22 area of green colour Commented by som(math1967) last updated on 07/Oct/22 $$\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\:?…
Question Number 111799 by ajfour last updated on 05/Sep/20 Commented by mr W last updated on 05/Sep/20 $${sir},\:{not}\:{possible}:\:{A}={B}={C} \\ $$ Commented by ajfour last updated…
Question Number 111734 by Aina Samuel Temidayo last updated on 04/Sep/20 $$\mathrm{A}\:\mathrm{chord}\:\mathrm{which}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perpendicular} \\ $$$$\mathrm{bisector}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{length}\:\mathrm{18cm}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circle},\:\mathrm{has}\:\mathrm{length}. \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated…