Question Number 83694 by naka3546 last updated on 05/Mar/20 Commented by naka3546 last updated on 05/Mar/20 $${DE}\:=\:{EF}\:=\:{FC}\:, \\ $$$${BG}\:=\:\mathrm{2}\:{CG}\:, \\ $$$${ABCD}\:\:{is}\:\:{a}\:\:{square}\:. \\ $$$$\frac{\left[\:{BMN}\:\right]}{\left[\:{ABCD}\:\right]}\:\:=\:\:? \\ $$$$\left(\:{without}\:\:{Trigonometry}\:\:{or}\:{Menelause}\:\right)…
Question Number 83690 by jagoll last updated on 05/Mar/20 $$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{\pi{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\right)}{{x}+\mathrm{1}}\:=\:? \\ $$ Commented by mathmax by abdo last updated on 05/Mar/20 $${let}\:{f}\left({x}\right)=\frac{{tan}\left(\frac{\pi{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\right)}{{x}+\mathrm{1}}\:\Rightarrow{f}\left({x}\right)=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}×\frac{{x}+\frac{\mathrm{1}}{\pi}}{{x}+\mathrm{1}}\right)}{{x}+\mathrm{1}} \\ $$$$=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}×\left(\frac{{x}+\mathrm{1}\:+\frac{\mathrm{1}}{\pi}−\mathrm{1}}{{x}+\mathrm{1}}\right)\right)}{{x}+\mathrm{1}}\:=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}−\pi}{\pi\left({x}+\mathrm{1}\right)}\right)\right.}{{x}+\mathrm{1}}…
Question Number 83691 by niroj last updated on 05/Mar/20 $$\:\mathrm{evaluate}: \\ $$$$\:\:\:\int\:\frac{\:\boldsymbol{\mathrm{dx}}}{\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{b}}\:\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}} \\ $$ Commented by mathmax by abdo last updated on 05/Mar/20 $${we}\:{use}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\Rightarrow \\…
Question Number 18153 by aplus last updated on 16/Jul/17 Commented by prakash jain last updated on 16/Jul/17 $$\mathrm{4}^{\mathrm{2}{x}} =\mathrm{2}×\mathrm{500}^{\mathrm{500}} \\ $$$$\mathrm{2}{x}\mathrm{log}\:\mathrm{4}=\mathrm{log}\:\mathrm{2}+\mathrm{500log}\:\mathrm{500} \\ $$$$\mathrm{3}=\frac{\mathrm{log}\:\mathrm{2}+\mathrm{500log}\:\mathrm{500}}{\mathrm{2log}\:\mathrm{4}}\approx\mathrm{1120}.\mathrm{973} \\ $$…
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Question Number 18149 by b.e.h.i.8.3.417@gmail.com last updated on 15/Jul/17 Commented by mrW1 last updated on 16/Jul/17 $$\mathrm{an}\:\mathrm{other}\:\mathrm{try}: \\ $$$$\mathrm{let}\:\mathrm{u}=\mathrm{xyz} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{xyz}=\mathrm{17x} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{17x}+\mathrm{u}=\mathrm{0}…
Question Number 18146 by mondodotto@gmail.com last updated on 15/Jul/17 Commented by prakash jain last updated on 16/Jul/17 $$\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{1} \\ $$$$\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \mathrm{3}{x}+\mathrm{9}{x}^{\mathrm{3}} +{x}}{\left(\mathrm{9}{x}^{\mathrm{2}}…
Question Number 149219 by mathdanisur last updated on 03/Aug/21 Answered by mindispower last updated on 03/Aug/21 $$\:\:^{\mathrm{4}} \sqrt{\varphi}={x} \\ $$$$\Leftrightarrow\frac{\mathrm{2}}{{x}}+{x}^{\mathrm{2}} <\mathrm{1}+\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +{x}^{\mathrm{3}} }{\mathrm{2}} \\…
Question Number 83680 by Power last updated on 05/Mar/20 Commented by john santu last updated on 05/Mar/20 $$\mathrm{considering}\:\mathrm{A}+\mathrm{B}\:=\:\mathrm{30}^{\mathrm{o}} \\ $$$$\mathrm{B}\:=\:\mathrm{30}^{\mathrm{o}} −\mathrm{A}\:\Rightarrow\:\mathrm{tan}\:\mathrm{B}\:=\:\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{tan}\:\mathrm{A}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}\:\mathrm{A}} \\ $$$$\mathrm{tan}\:\mathrm{B}\:=\:\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{A}}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}} \\ $$$$\sqrt{\mathrm{3}}\:+\:\mathrm{tan}\:\mathrm{B}\:=\:\sqrt{\mathrm{3}}+\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{A}}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}}…
Question Number 149212 by mathdanisur last updated on 03/Aug/21 $${if}\:\:\:{x};{y};{z}>\mathrm{0}\:\:\:{and}\:\:\:{xyz}=\mathrm{1}\:\:\:{prove}\:{that}: \\ $$$$\frac{{x}^{\mathrm{4}} }{{x}+{yz}}\:+\:\frac{{y}^{\mathrm{4}} }{{y}+{zx}}\:+\:\frac{{z}^{\mathrm{4}} }{{z}+{xy}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Answered by dumitrel last updated on 03/Aug/21 Commented…