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Question-197989

Question Number 197989 by mr W last updated on 07/Oct/23 Commented by mr W last updated on 07/Oct/23 $$\left[\underline{{old}\:{question}\:{from}\:{ajfour}\:{sir}}\right] \\ $$$$\mathrm{3}\:{circles}\:{with}\:{radii}\:{p},\:{q},\:{r}\:{respectively} \\ $$$${touch}\:{each}\:{other}\:{as}\:{shown}.\:{find}\:{the}\: \\ $$$${maximum}\:{area}\:{of}\:\:{triangle}\:\Delta{ABC}…

Question-198018

Question Number 198018 by obia last updated on 07/Oct/23 Answered by a.lgnaoui last updated on 08/Oct/23 $$\boldsymbol{\mathrm{Z}}=\sqrt{\mathrm{3}}\:+\mathrm{1}+\:\boldsymbol{\mathrm{i}}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right) \\ $$$$ \\ $$$$\left.\bullet\mathrm{1}\right):\:\:\boldsymbol{\mathrm{Z}}^{\mathrm{2}} =\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\:−\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\:\right)+\mathrm{2}\boldsymbol{\mathrm{i}}\left(\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{3}}\:−\:\mathrm{1}\right) \\ $$$$\:\:\:=\mathrm{4}\sqrt{\mathrm{3}}\:\:+\mathrm{4}\boldsymbol{\mathrm{i}}\:\:=\mathrm{4}\left(\sqrt{\mathrm{3}}\:+\boldsymbol{\mathrm{i}}\right) \\…

Question-197967

Question Number 197967 by Blackpanther last updated on 06/Oct/23 Answered by mr W last updated on 06/Oct/23 $${f}\left({x}\right)={a}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{10}\right) \\ $$$$\frac{\mathrm{20}}{\mathrm{3}}={a}\left(\mathrm{0}+\mathrm{1}\right)\left(\mathrm{0}−\mathrm{10}\right) \\ $$$$\Rightarrow{a}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${at}\:{x}=\frac{−\mathrm{1}+\mathrm{10}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}}: \\…

determiner-le-total-de-nombres-de-5-chiffres-comprises-entre-10000-et-50000-divisibles-simultanement-par-5-et-9-sans-utiliser-les-formules-d-arrangement-et-de-combinaison-

Question Number 197960 by a.lgnaoui last updated on 06/Oct/23 $$\mathrm{determiner}\:\mathrm{le}\:\mathrm{total}\:\mathrm{de}\:\mathrm{nombres}\:\mathrm{de}\: \\ $$$$\mathrm{5}\:\mathrm{chiffres}\:\mathrm{comprises}\:\mathrm{entre}\:\mathrm{10000}\:\mathrm{et}\: \\ $$$$\mathrm{50000}\:\:\mathrm{divisibles}\:\mathrm{simultanement}\:\mathrm{par} \\ $$$$\mathrm{5}\:\mathrm{et}\:\mathrm{9}\:\:\: \\ $$$$\left(\mathrm{sans}\:\mathrm{utiliser}\:\mathrm{les}\:\mathrm{formules}\:\mathrm{d}\:\mathrm{arrangement}\right. \\ $$$$\left.\mathrm{et}\:\mathrm{de}\:\mathrm{combinaison}\right) \\ $$$$ \\ $$ Answered…

Let-x-y-z-gt-0-x-y-z-3-Prove-That-1-x-2-2x-1-z-2-2z-3-1-y-2-y-9-x-y-z-24-1-3-3-17-3-3-

Question Number 197950 by York12 last updated on 05/Oct/23 $${Let}\:{x},{y},{z}>\mathrm{0}\:,\:{x}+{y}+{z}=\mathrm{3}\:{Prove}\:{That}\:: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}+\frac{\mathrm{1}}{\:\sqrt{{z}^{\mathrm{2}} +\mathrm{2}{z}}}+\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{{y}+\mathrm{2}}−\frac{{y}}{\mathrm{9}}\right)+\frac{\sqrt[{\mathrm{3}}]{\sqrt{{x}}+\sqrt{{y}}+\sqrt{{z}}+\mathrm{24}}}{\:\sqrt{\mathrm{3}}}\geqslant\frac{\mathrm{17}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com