Question Number 193149 by Mingma last updated on 04/Jun/23 Answered by ajfour last updated on 05/Jun/23 $$\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{p}^{\mathrm{2}} ={A} \\ $$$${p}={k}\sqrt{{A}}\:\:\:\:{where}\:\:\:{k}^{\mathrm{2}} =\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$${q}={k}\sqrt{{B}} \\ $$$${B}=\mathrm{9}{A}…
Question Number 193137 by Mastermind last updated on 04/Jun/23 $$\left.\mathrm{1}\right)\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{a}+\mathrm{b}+\mathrm{c}\mid\geqslant\mid\mathrm{a}\mid−\mid\mathrm{b}\mid−\mid\mathrm{c}\mid \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Find}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}\:\mathrm{that}\:\mathrm{satify}\:\mathrm{the}\:\mathrm{follow}− \\ $$$$\mathrm{ing}\:\mathrm{inequalities}\: \\ $$$$\left.\mathrm{i}\right)\:\mid\mathrm{x}^{\mathrm{2}} −\mathrm{4}\mid<\mathrm{5} \\ $$$$\left.\mathrm{ii}\right)\:\mid\mathrm{x}\mid+\mid\mathrm{x}+\mathrm{2}\mid<\mathrm{5} \\ $$$$…
Question Number 193139 by Mastermind last updated on 04/Jun/23 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{inequalities}: \\ $$$$\mid\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mid>\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$ Answered by Skabetix last updated on 04/Jun/23…
Question Number 193138 by Mastermind last updated on 04/Jun/23 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b}\in\mathbb{R} \\ $$$$\left.\mathrm{i}\right)\:\mathrm{ab}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{ii}\right)\:\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{iii}\right)\:\sqrt{\mathrm{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}+\mathrm{b}\right),\:\mathrm{for}\:\mathrm{a},\mathrm{b}\geqslant\mathrm{0}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{have}\:\mathrm{square}\:\mathrm{roots}. \\ $$$$…
Question Number 193130 by Frix last updated on 04/Jun/23 $$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\mathrm{2}} −{c}=\sqrt{{c}−{x}} \\ $$ Commented by aba last updated on 04/Jun/23 $$\mathrm{t}=\mathrm{c}−\mathrm{x}\geqslant\mathrm{0}\: \\ $$$$\mathrm{x}^{\mathrm{2}}…
Question Number 193117 by mustafazaheen last updated on 04/Jun/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cosx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$ Answered by Subhi last updated on 04/Jun/23 $${y}\:=\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left({cosx}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$${ln}\left({y}\right)\:=\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{ln}\left({cosx}\right)}{{x}}…
Question Number 193116 by Mastermind last updated on 04/Jun/23 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\in\:\mathbb{R}\:\mathrm{with} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\geqslant\:\mathrm{0}\: \\ $$$$\left.\mathrm{1}\right)\:\sqrt{\mathrm{ab}}\sqrt{\mathrm{cd}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{2}\right)\:\left(\mathrm{abcd}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right) \\ $$$$ \\ $$$$\mathrm{Help}!…
Question Number 193109 by Mastermind last updated on 04/Jun/23 Answered by math1234 last updated on 04/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193111 by cortano12 last updated on 04/Jun/23 $$\:\:\:\:\:\:\downharpoonleft\underline{\:} \\ $$ Answered by math1234 last updated on 04/Jun/23 Answered by horsebrand11 last updated on…
Question Number 193105 by Mingma last updated on 04/Jun/23 Commented by Mingma last updated on 04/Jun/23 Find x Terms of Service Privacy Policy Contact: info@tinkutara.com