Question Number 227370 by Math1 last updated on 18/Jan/26 $$\frac{\mathrm{16}\centerdot\mathrm{3}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} \:+\:\mathrm{16}\centerdot\mathrm{3}^{\boldsymbol{\mathrm{x}}} }{\mathrm{4}\centerdot\mathrm{3}^{\boldsymbol{\mathrm{x}}+\mathrm{1}} \:−\:\mathrm{2}\centerdot\mathrm{3}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} }\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} }}{\mathrm{17}}\:\:\:\:\:\mathrm{find}:\:\:\mathrm{x}=? \\ $$ Answered by Kassista last updated on 18/Jan/26 $$…
Question Number 227371 by Math1 last updated on 18/Jan/26 $$\mathrm{3}^{\mathrm{444}} \:\:+\:\:\mathrm{4}^{\mathrm{333}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{dividing}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{by}\:\mathrm{7} \\ $$ Answered by A5T last updated on 19/Jan/26 $$\phi\left(\mathrm{7}\right)=\mathrm{6}…
Question Number 227361 by Hanuda354 last updated on 18/Jan/26 Commented by Hanuda354 last updated on 18/Jan/26 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{area},\:\mathrm{where} \\ $$$$\mathrm{A}\:=\:\mathrm{B}\:=\:\mathrm{24}.\mathrm{8}\:\mathrm{square}\:\mathrm{units}. \\ $$$$\left(\mathrm{A},\:\mathrm{B}\:\mathrm{are}\:\mathrm{shaded}\:\mathrm{regions}\right) \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{square}\:\mathrm{area}. \\ $$…
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Question Number 227341 by fantastic2 last updated on 17/Jan/26 $${find}\:{the}\:{surface}\:{area}\:{of}\:{a}\:{spherical} \\ $$$${cap} \\ $$ Commented by Kassista last updated on 17/Jan/26 $$ \\ $$$${are}\:{you}\:{simply}\:{asking}\:{to}\:{prove}\:{the}\:{surface}\:{area} \\…
Question Number 227353 by fantastic2 last updated on 17/Jan/26 Answered by mr W last updated on 19/Jan/26 Commented by mr W last updated on 19/Jan/26…
Question Number 227329 by Lara2440 last updated on 17/Jan/26 $$\left.\mathrm{1}\right)\:\mathrm{Does}\:\mathrm{half}\:\mathrm{open}\:\mathrm{interval}\:{A}=\left[\mathrm{0},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{Compact}\:\mathrm{Space}\:? \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Prove}\:\mathrm{for}\:\mathrm{a}\:\:\mathrm{Compact}\:\mathrm{Space}\:{X}_{{k}} \:\: \\ $$$$\mathrm{Product}\:\mathrm{Space}\:{X}=\underset{{k}} {\prod}\:\left\{{X}_{{k}} ^{\:} \:;\:{k}\in{I}\right\}\:\:\mathrm{also}\:\mathrm{Compact}\:\mathrm{Space} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{a}\:\mathrm{Compact}\:\mathrm{Space}\:{X}\:\mathrm{and}\:\mathrm{Continuous}\:\mathrm{function}\:{f} \\ $$$$\mathrm{if}\:\:{f}\:\:\mathrm{satisfy}\:\:{f};{X}\rightarrow{Y}\:,\mathrm{Image}\:{Y}\:\mathrm{also}\:\mathrm{Compact}\:\mathrm{Space} \\ $$ Terms…
Question Number 227346 by Spillover last updated on 17/Jan/26 Answered by Kassista last updated on 18/Jan/26 $$ \\ $$$${Note}\:{that}: \\ $$$$\lfloor{x}\rfloor=\:{n},\:{if}\:{n}\leqslant{x}<{n}+\mathrm{1}\:\left({n}\in\mathbb{Z}\right) \\ $$$${similarly}: \\ $$$$\lceil{x}\rceil={n}+\mathrm{1},\:{n}\leqslant{x}<{n}+\mathrm{1}\:\left({n}\in\mathbb{Z}\right)…
Question Number 227325 by hardmath last updated on 16/Jan/26 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{any}\:\mathrm{acute}\:\bigtriangleup\mathrm{ABC}\: \\ $$$$\mathrm{if}\:\mathrm{I}\:\mathrm{is}\:\mathrm{the}\:\mathrm{in}-\mathrm{center}\:\mathrm{and}\:\mathrm{H}\:\mathrm{is}\:\mathrm{the}\:\mathrm{ortho}-\mathrm{center} \\ $$$$\mathrm{then}: \\ $$$$\frac{\mathrm{1}}{\mathrm{IA}}\:+\:\frac{\mathrm{1}}{\mathrm{IB}}\:+\:\frac{\mathrm{1}}{\mathrm{IC}}\:\:\leqslant\:\:\frac{\mathrm{1}}{\mathrm{HA}}\:+\:\frac{\mathrm{1}}{\mathrm{HB}}\:+\:\frac{\mathrm{1}}{\mathrm{HC}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 227326 by gregori last updated on 16/Jan/26 $$\:\:\:\mid\:{x}+\mathrm{1}\:\mid\:+\:\mid\:{x}\:\mid\:+\:\mid\:{x}−\mathrm{3}\mid\:>\:\mathrm{8}\:\: \\ $$ Answered by Kassista last updated on 16/Jan/26 $$ \\ $$$${if}\:{x}\geqslant\mathrm{3}: \\ $$$${x}+\mathrm{1}+{x}+{x}−\mathrm{3}>\mathrm{8}\:\Leftrightarrow\:\mathrm{3}{x}−\mathrm{2}>\mathrm{8},\:\mathrm{3}{x}>\mathrm{10},\:{x}>\frac{\mathrm{10}}{\mathrm{3}}…\left(\mathrm{1}\right) \\…