Question Number 61322 by Tawa1 last updated on 31/May/19 Answered by MJS last updated on 02/Jun/19 $$\mathrm{the}\:\mathrm{cuboid}\:\mathrm{with}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{volume}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{given}\:\mathrm{surface}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube} \\ $$$$ \\ $$$${S}\mathrm{urface}=\mathrm{2}{a}^{\mathrm{2}} +\mathrm{4}{ab}=\mathrm{216}\:\Rightarrow\:{b}=\frac{\mathrm{54}}{{a}}−\frac{{a}}{\mathrm{2}} \\…
Question Number 61303 by mr W last updated on 31/May/19 Commented by mr W last updated on 31/May/19 $${A}\:{small}\:{sphere}\:\left({say}\:{the}\:{moon}\right)\:{moves} \\ $$$${around}\:{a}\:{big}\:{sphere}\:\left({say}\:{the}\:{earth}\right)\:{in} \\ $$$${a}\:{circular}\:{track}.\:{A}\:{point}\:{source}\:\left({say}\right. \\ $$$$\left.{the}\:{sun}\right)\:{is}\:{at}\:{a}\:{distance}\:{e}\:{from}\:{the}…
Question Number 192325 by cherokeesay last updated on 14/May/23 Answered by a.lgnaoui last updated on 14/May/23 $$\boldsymbol{\mathrm{s}}\mathrm{urface}\:\boldsymbol{\mathrm{S}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{\mathrm{AB}}×\boldsymbol{\mathrm{BF}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{CD}}×\left(\boldsymbol{\mathrm{BD}}−\boldsymbol{\mathrm{DF}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{S}}=\mathrm{2}\boldsymbol{\mathrm{CD}} \\ $$$$\boldsymbol{\mathrm{C}}\mathrm{alcul}\:\:\boldsymbol{\mathrm{CD}} \\ $$$$\bigtriangleup\mathrm{ACE}\:\:\measuredangle\mathrm{EAC}=\measuredangle\mathrm{DEF}=\boldsymbol{\theta} \\ $$$$\boldsymbol{\mathrm{C}}\mathrm{E}=\mathrm{AEsin}\:\theta\:\:\:\mathrm{cos}\:\theta=\frac{\mathrm{5}}{\mathrm{AE}}\Rightarrow\mathrm{AE}=\frac{\mathrm{5}}{\mathrm{cos}\:\theta}…
Question Number 61235 by ajfour last updated on 30/May/19 Commented by ajfour last updated on 30/May/19 $${Find}\:{h}\:{and}\:{R}\:{of}\:{maximum}\:{volume} \\ $$$$\:{inscribed}\:{cylinder}\:{with}\:{its}\:{base}\:{on} \\ $$$${face}\:{ABC}\:{of}\:{a}\:{general}\:{pyramid}. \\ $$ Commented by…
Question Number 192288 by universe last updated on 14/May/23 Answered by mahdipoor last updated on 14/May/23 $${get}\:\angle{BCG}={a}\:\:\:\:\:,\:{BC}={b} \\ $$$${GD}={GB}+{BD}={BC}.{tana}+{BC}.{tan}\left(\mathrm{90}−{a}\right) \\ $$$$={b}\left({tana}+{cota}\right) \\ $$$${FD}={GD}.{tana}={b}\left({tan}^{\mathrm{2}} {a}+\mathrm{1}\right)={b}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {a}}\right)…
Question Number 61186 by Tawa1 last updated on 30/May/19 Commented by Prithwish sen last updated on 30/May/19 $$\mathrm{Area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shadded}\:\mathrm{portion}\:=\mathrm{Area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{incircle}\: \\ $$$$−\mathrm{Area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quarter}\:\mathrm{circle}\:+\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{Area}\:\mathrm{of}\:\mathrm{the}\:\right. \\ $$$$\left.\mathrm{square}\:−\mathrm{Area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{incircle}\right) \\ $$$$=\pi\mathrm{5}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{10}\right)^{\mathrm{2}}…
Question Number 61169 by Tawa1 last updated on 29/May/19 Commented by Tawa1 last updated on 29/May/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Black} \\ $$ Answered by mr W last updated…
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Question Number 126609 by abdullahquwatan last updated on 22/Dec/20 $$\mathrm{cube}\:\mathrm{ABCD}.\mathrm{EFGH}\:{side}\:\mathrm{4}\:{cm}\:{point}\:{P}\:{is}\:{center}\:{BF}\:{and}\:{Q}\:{center}\:{AD}.\:{distance}\:{E}\:{to}\:{PQG} \\ $$ Answered by liberty last updated on 22/Dec/20 $${the}\:{equation}\:{of}\:{plane}\:{passes}\:{trought} \\ $$$${P}\left(\mathrm{4},\mathrm{4},\mathrm{2}\right),{Q}\left(\mathrm{2},\mathrm{0},\mathrm{0}\right),{G}\left(\mathrm{0},\mathrm{4},\mathrm{4}\right) \\ $$$$\Rightarrow\:\begin{vmatrix}{{x}−\mathrm{2}\:\:\:\:\:\:\:{y}\:\:\:\:\:\:\:\:\:\:{z}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{4}}\\{\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}=\:\mathrm{0} \\…