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Category: Algebra

16-3-x-1-16-3-x-4-3-x-1-2-3-x-1-2-x-1-1-3-17-find-x-

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-227346

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)…

Prove-that-in-any-acute-ABC-if-I-is-the-in-center-and-H-is-the-ortho-center-then-1-IA-1-IB-1-IC-1-HA-1-HB-1-HC-

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

tg-15-ctg-5-

Question Number 227312 by hardmath last updated on 13/Jan/26 $$\mathrm{tg}\left(\mathrm{15}\right)\:+\:\mathrm{ctg}\left(\mathrm{5}\right)\:=\:? \\ $$ Answered by Kassista last updated on 13/Jan/26 $$ \\ $$$${tg}\left(\mathrm{3}\theta\right)\:=\:\frac{\mathrm{3}{tg}\left(\theta\right)−{tg}^{\mathrm{3}} \left(\theta\right)}{\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\theta\right)}\:\therefore\:{tg}\left(\mathrm{15}\right)=\frac{\mathrm{3}{tg}\left(\mathrm{5}\right)−{tg}^{\mathrm{3}} \left(\mathrm{5}\right)}{\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}}…