Question Number 110673 by Aina Samuel Temidayo last updated on 30/Aug/20 $$ \\ $$$$\mathrm{Let}\:\mathrm{k}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\sqrt{\mathrm{x}−\mathrm{3}}+\sqrt{\mathrm{6}−\mathrm{x}}\geqslant\mathrm{k}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{solution}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{k}. \\ $$ Answered by Her_Majesty…
Question Number 176206 by mathlove last updated on 15/Sep/22 Answered by TheHoneyCat last updated on 18/Sep/22 $$\mathrm{a}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{1}^{\mathrm{1}} .\left(\left(\mathrm{2}^{\mathrm{1}} \right)^{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{1}.\left(\mathrm{2}^{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}}…
Question Number 176189 by Linton last updated on 14/Sep/22 $${solve}\:{for}\:{x} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} \\ $$$${show}\:{working} \\ $$ Answered by BaliramKumar last updated on 14/Sep/22 $$\left(\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}}…
Question Number 176159 by peter frank last updated on 14/Sep/22 Answered by adhigenz last updated on 14/Sep/22 $$\mathrm{Let}\:{x}\:=\:\mathrm{cos}\:\alpha\:+\:{i}.\mathrm{sin}\:\alpha\:=\:{e}^{{i}.\alpha} \:\mathrm{and}\:{y}\:=\:\mathrm{cos}\:\beta\:+\:{i}.\mathrm{sin}\:\beta\:=\:{e}^{{i}.\beta} \\ $$$$\left(\mathrm{a}\right).\:{x}^{{m}} {y}^{{n}} \:+\:\frac{\mathrm{1}}{{x}^{{m}} {y}^{{n}} }\:=\:{e}^{{i}.{m}\alpha}…
Question Number 110620 by peter frank last updated on 29/Aug/20 Answered by Dwaipayan Shikari last updated on 29/Aug/20 $${f}\left({x}\right)={tan}^{−\mathrm{1}} {x} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \\…
Question Number 176155 by adhigenz last updated on 14/Sep/22 $$\mathrm{Find}\:\mathrm{how}\:\mathrm{many}\:\mathrm{distinct}\:\mathrm{integers}\:\mathrm{are}\:\mathrm{there}\:\mathrm{in}\:\mathrm{this}\:\mathrm{sequence}: \\ $$$$\lfloor\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{1}}{\mathrm{100}}\rfloor,\:\lfloor\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{2}}{\mathrm{100}}\rfloor,\:\lfloor\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{3}}{\mathrm{100}}\rfloor,\:…,\:\lfloor\frac{\mathrm{100}^{\mathrm{2}} +\mathrm{100}}{\mathrm{100}}\rfloor \\ $$$$\mathrm{where}\:\lfloor{x}\rfloor\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:{x} \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 110608 by Engr_Jidda last updated on 29/Aug/20 $$\sqrt{{x}}\:+\mathrm{1}=\mathrm{0} \\ $$$${Solution} \\ $$$$\sqrt{{x}\:}\:=\:−\mathrm{1} \\ $$$${recalled}\:{that}\:\varrho^{{i}\Pi} =\:−\mathrm{1} \\ $$$$\sqrt{{x}}\:=\varrho^{{i}\Pi} \\ $$$${x}=\left(\varrho^{{i}\Pi} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\varrho^{\mathrm{2}{i}\Pi} \\…
Question Number 176141 by Linton last updated on 13/Sep/22 $${which}\:{is}\:{greater}\:? \\ $$$$\mathrm{4}^{\sqrt{\mathrm{2}}} \:{or}\:\mathrm{3}^{\sqrt{\mathrm{3}}} \\ $$ Commented by Linton last updated on 13/Sep/22 $${without}\:{calculator}\:{show}\:{working} \\ $$…
Question Number 110596 by Aina Samuel Temidayo last updated on 29/Aug/20 $$\mathrm{Three}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{satisfying} \\ $$$$\mathrm{ab}+\mathrm{c}=\mathrm{10},\mathrm{bc}+\mathrm{a}=\mathrm{11},\mathrm{ca}+\mathrm{b}=\mathrm{14}.\:\mathrm{Find} \\ $$$$\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{b}−\mathrm{c}\right)\left(\mathrm{c}−\mathrm{a}\right)\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{b}−\mathrm{1}\right)\left(\mathrm{c}−\mathrm{1}\right) \\ $$ Commented by Her_Majesty last updated on 29/Aug/20…
Question Number 176128 by Shrinava last updated on 13/Sep/22 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC}\:,\:\mathrm{cot}\:\frac{\mathrm{A}}{\mathrm{2}}\:,\:\mathrm{cot}\:\frac{\mathrm{B}}{\mathrm{2}}\:,\:\mathrm{cot}\:\frac{\mathrm{C}}{\mathrm{2}}\:\in\:\mathrm{Q} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\left(\underset{\boldsymbol{\mathrm{cyc}}} {\prod}\:\mathrm{sin}\:\frac{\mathrm{A}}{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} +\:\left(\underset{\boldsymbol{\mathrm{cyc}}} {\prod}\:\mathrm{cos}\:\frac{\mathrm{A}}{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} \in\:\mathrm{Q}\:,\:\forall\mathrm{n}\in\mathbb{N} \\ $$ Commented by Rasheed.Sindhi last updated…