Question Number 17440 by virus last updated on 05/Jul/17 $$\mathrm{3}{x}−\mathrm{4}{y}=\mathrm{0},\mathrm{4}{y}−\mathrm{5}{z}=\mathrm{0},\mathrm{5}{z}−\mathrm{3}{x}=\mathrm{0} \\ $$$${then}\:{x},{y},{z}\:{is}\:{AP},{GP},{HP},{AGP}?????? \\ $$ Answered by sushmitak last updated on 06/Jul/17 $$\frac{{x}}{{y}}=\frac{\mathrm{4}}{\mathrm{3}}={k},\frac{{y}}{{z}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${x}=\mathrm{4}{k} \\…
Question Number 17373 by mrW1 last updated on 04/Jul/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex} \\ $$$$\mathrm{quadrilateral}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{the}\:\mathrm{4}\:\mathrm{vertices}\:\mathrm{is}\:\mathrm{minimal}. \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex} \\ $$$$\mathrm{quadrilateral}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{the}\:\mathrm{4}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{minimal}. \\ $$ Answered…
Question Number 82901 by Power last updated on 25/Feb/20 Answered by mr W last updated on 26/Feb/20 Commented by Power last updated on 26/Feb/20 $$\mathrm{sir}\:\:\mathrm{a}:\mathrm{b}:\mathrm{c}=?…
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Question Number 82878 by mr W last updated on 25/Feb/20 Commented by behi83417@gmail.com last updated on 25/Feb/20 $$\mathrm{Beautiful}\:\mathrm{quistion}.\mathrm{waiting}\:\mathrm{for}\:\mathrm{answer}… \\ $$ Commented by ajfour last updated…
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Question Number 17273 by VEGAMIND last updated on 03/Jul/17 $$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{intersection}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ABC}}\:\boldsymbol{\mathrm{triangle}} \\ $$$$\boldsymbol{\mathrm{median}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{G}}\:\boldsymbol{\mathrm{point}}.\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{corner}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{BGC}}\:\boldsymbol{\mathrm{is}}\:\mathrm{90}°.\:\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{AG}}\:\boldsymbol{\mathrm{cut}}\: \\ $$$$\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{is}}\:\mathrm{12}\:\boldsymbol{\mathrm{cm}},\:\boldsymbol{\mathrm{locate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{BC}}\:\boldsymbol{\mathrm{side}}. \\ $$ Commented by 1234Hello last updated on 03/Jul/17…
Question Number 17177 by tawa tawa last updated on 01/Jul/17 Answered by Tinkutara last updated on 02/Jul/17 $$\mathrm{Tangential}\:\mathrm{acceleration}\:\mathrm{will}\:\mathrm{be}\:\mathrm{0}. \\ $$$$\mathrm{Frequency},\:\nu\:=\:\frac{\mathrm{300}}{\mathrm{60}}\:=\:\mathrm{5}\:\mathrm{Hz} \\ $$$$\mathrm{Angular}\:\mathrm{velocity},\:\omega\:=\:\mathrm{2}\pi×\mathrm{5}\:=\:\mathrm{10}\pi\:\mathrm{rad}/\mathrm{s} \\ $$$$\mathrm{Centripetal}\:\mathrm{acceleration}\:\mathrm{cannot}\:\mathrm{be} \\…
Question Number 17158 by Tinkutara last updated on 01/Jul/17 $$\mathrm{Please}\:\mathrm{solve}\:\mathrm{Q}.\:\mathrm{16069}.\:\mathrm{Ask}\:\mathrm{from}\:\mathrm{me}\:\mathrm{the} \\ $$$$\mathrm{solution}\:\mathrm{if}\:\mathrm{needed}\:\mathrm{and}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{it}. \\ $$ Commented by mrW1 last updated on 01/Jul/17 $$\mathrm{one}\:\mathrm{can}\:\mathrm{prove} \\ $$$$\mathrm{case}\:\mathrm{1}:\:\mathrm{AB}\:\mathrm{not}\://\:\mathrm{to}\:\mathrm{CD}\:\mathrm{and}\:\mathrm{AD}\:\mathrm{not}\://\:\mathrm{to}\:\mathrm{BC}: \\…
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