Question Number 88961 by Zainal Arifin last updated on 14/Apr/20 $$\mathrm{One}\:\mathrm{care}\:\mathrm{travels}\:\mathrm{due}\:\mathrm{east}\:\mathrm{at}\:\mathrm{40}\:\mathrm{km}/\mathrm{h}. \\ $$$$\mathrm{and}\:\mathrm{second}\:\mathrm{car}\:\mathrm{travels}\:\mathrm{north}\:\mathrm{at}\: \\ $$$$\mathrm{40}\:\mathrm{km}/\mathrm{h}.\:\mathrm{are}\:\mathrm{their}\:\mathrm{velocities}\:\mathrm{equal}? \\ $$ Commented by ajfour last updated on 14/Apr/20 $${No},\:{since}\:{directions}\:{are}\:{not}\:{same}.…
Question Number 154493 by mathdanisur last updated on 18/Sep/21 $$\mathrm{If}\:\:\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{n}\in\mathbb{N}^{+} \:\:\mathrm{then}: \\ $$$$\frac{\mathrm{a}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \:+\:\mathrm{b}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \:+\:\mathrm{c}^{\mathrm{2}\boldsymbol{\mathrm{n}}} }{\mathrm{a}^{\boldsymbol{\mathrm{n}}} \mathrm{b}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{b}^{\boldsymbol{\mathrm{n}}} \mathrm{c}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{c}^{\boldsymbol{\mathrm{n}}} \mathrm{a}^{\boldsymbol{\mathrm{n}}} }\:\geqslant\:\frac{\sqrt{\mathrm{3}\centerdot\left(\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \right)}}{\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}}…
Question Number 154495 by mathdanisur last updated on 18/Sep/21 $$\mathrm{if}\:\:\mathrm{S}_{\boldsymbol{\mathrm{n}}} \left(\mathrm{t}\right)\:=\:\mathrm{n}^{\mathrm{1}-\boldsymbol{\mathrm{t}}} \:\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}\boldsymbol{\mathrm{t}}} }{\left(\sqrt[{\boldsymbol{\mathrm{n}}+\mathrm{1}}]{\left(\mathrm{n}+\mathrm{1}\right)!}\right)^{\boldsymbol{\mathrm{t}}} }\:-\:\frac{\mathrm{n}^{\mathrm{2}\boldsymbol{\mathrm{t}}} }{\left(\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\right)^{\boldsymbol{\mathrm{t}}} }\right) \\ $$$$\mathrm{with}\:\:\mathrm{t}>\mathrm{0} \\ $$$$\mathrm{then}\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}S}_{\boldsymbol{\mathrm{n}}} \left(\mathrm{t}\right)\:=\:\mathrm{te}^{\boldsymbol{\mathrm{t}}} \\ $$ Answered…
Question Number 23419 by mondodotto@gmail.com last updated on 30/Oct/17 Answered by $@ty@m last updated on 31/Oct/17 $${Solution}\:\left({a}\right) \\ $$$${LHS}=\mathrm{cos}\:{A}+\mathrm{cos}\:{B}+\mathrm{cos}\:{C} \\ $$$$=\mathrm{2cos}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{cos}\:{C} \\ $$$$=\mathrm{2cos}\:\frac{\pi−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{cos}\:{C} \\ $$$$=\mathrm{2sin}\:\frac{{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}−\mathrm{2sin}\:^{\mathrm{2}}…
Question Number 88955 by M±th+et£s last updated on 14/Apr/20 $${hello}\: \\ $$$${floor}\:{function} \\ $$$$\int_{{a}} ^{{b}} \lfloor{x}\rfloor\:\:{dx}\:\:\:\:\:\:\:\:\:\:{a},{b}\in{z}\:\:\:{and}\:{b}>{a} \\ $$$$=\int_{\mathrm{0}} ^{{b}} \lfloor{x}\rfloor\:{dx}\:−\int_{\mathrm{0}} ^{{a}} \lfloor{x}\rfloor\:{dx}\:=\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}\:….\left(\mathrm{1}\right) \\…
Question Number 23418 by tapan das last updated on 30/Oct/17 $$\int\mathrm{sec}\:^{\mathrm{2}} \sqrt{\mathrm{x}}\:/\sqrt{\mathrm{x}}\:\mathrm{dx} \\ $$ Answered by $@ty@m last updated on 30/Oct/17 $${Assume}\:\sqrt{{x}}={t}\:\&\:{proceed}. \\ $$$${Ans}\:\mathrm{2tan}\:\sqrt{{x}}+{C} \\…
Question Number 88951 by M±th+et£s last updated on 14/Apr/20 $$\int_{{a}} ^{{b}} \lceil{x}\rceil\:{dx}=? \\ $$$${a},{b}\in{R}\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\lceil..\rceil\:{is}\:{ceil}\: \\ $$ Answered by mr W last…
Question Number 23411 by ajfour last updated on 29/Oct/17 Commented by ajfour last updated on 29/Oct/17 $${Place}\:\mathrm{8}\:{queens}\:{on}\:{the}\:{chessboard} \\ $$$${so}\:{that}\:{there}\:{be}\:{no}\:{threat}\:\left({none}\right. \\ $$$$\left.{come}\:{in}\:{each}\:{other}'{s}\:{way}\right). \\ $$$$ \\ $$…
Question Number 23409 by Tinkutara last updated on 29/Oct/17 $$\mathrm{Hybridisation}\:\mathrm{of}\:\mathrm{N}\:\mathrm{in}\:\mathrm{HNO}_{\mathrm{3}} \:\mathrm{is} \\ $$ Answered by Tinkutara last updated on 01/Nov/17 Terms of Service Privacy Policy…
Question Number 23408 by selestian last updated on 29/Oct/17 Answered by mrW1 last updated on 31/Oct/17 $$\mathrm{FE}=\frac{\mathrm{5}}{\mathrm{10}}×\mathrm{18}=\mathrm{9} \\ $$$$\mathrm{AC}=\frac{\mathrm{15}}{\mathrm{9}}×\left(\mathrm{5}+\mathrm{10}\right)=\mathrm{25} \\ $$$$\Rightarrow\mathrm{Answer}\:\left(\mathrm{b}\right) \\ $$ Terms of…