Question Number 127018 by ‘-E/9 last updated on 26/Dec/20 $${Vf}×\mathrm{2}={Vi}×\mathrm{2}+\mathrm{2}{a}\Delta{d} \\ $$$$\mathrm{0}=\mathrm{16}.\mathrm{5} \\ $$ Answered by physicstutes last updated on 26/Dec/20 $$\mathrm{you}\:\mathrm{should}\:\mathrm{write}\:\mathrm{it}\:\mathrm{this}\:\mathrm{way}. \\ $$$$\:\:{v}_{{f}} ^{\mathrm{2}}…
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Question Number 61449 by Tawa1 last updated on 02/Jun/19 Commented by Tawa1 last updated on 02/Jun/19 $$\frac{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} }\:\:=\:\:? \\ $$ Answered by perlman…
Question Number 61424 by Tawa1 last updated on 02/Jun/19 Answered by mr W last updated on 02/Jun/19 Commented by mr W last updated on 02/Jun/19…
Question Number 192469 by Mingma last updated on 18/May/23 Answered by a.lgnaoui last updated on 19/May/23 $$\bigtriangleup\mathrm{ABC}\:\:\mathrm{triangle}\:\mathrm{isocele}\:\:\mathrm{AB}=\mathrm{AC} \\ $$$$\mathrm{donc}\:\:\measuredangle\mathrm{B}=\measuredangle\mathrm{C}=\mathrm{90}−\frac{\mathrm{78}}{\mathrm{2}}=\mathrm{51}° \\ $$$$\:\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{78}}{\boldsymbol{\mathrm{BC}}}=\frac{\mathrm{sin}\:\mathrm{51}}{\boldsymbol{\mathrm{AC}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup\mathrm{ABD}\:\:\:\:\mathrm{AB}=\mathrm{BD}\:\:\:\measuredangle\mathrm{CAD}=\mathrm{A}_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:…
Question Number 61347 by Tawa1 last updated on 01/Jun/19 Commented by mr W last updated on 02/Jun/19 $${x}=\mathrm{20}°,\:{see}\:{also}\:{Q}\mathrm{43728}. \\ $$ Answered by peter frank last…
Question Number 61335 by necx1 last updated on 01/Jun/19 Answered by mr W last updated on 02/Jun/19 $$\mathrm{cos}\:\angle{A}=\frac{\mathrm{5}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}×\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{A}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\angle{B}=\frac{\mathrm{5}^{\mathrm{2}}…
Question Number 61322 by Tawa1 last updated on 31/May/19 Answered by MJS last updated on 02/Jun/19 $$\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}…