Question Number 77046 by ajfour last updated on 02/Jan/20 $${x}={R}^{\mathrm{2}} \sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{{R}^{\mathrm{2}} }}\: \\ $$ Answered by ajfour last updated on 02/Jan/20 Commented by necxxx…
Question Number 142582 by mathdanisur last updated on 02/Jun/21 $$\boldsymbol{{z}}^{\mathrm{4}} \:+\:\mathrm{12}\boldsymbol{{z}}\:+\:\mathrm{3}\:=\:\mathrm{0}\:\left(\boldsymbol{{z}}=?\right) \\ $$ Answered by ajfour last updated on 02/Jun/21 $$\left({z}^{\mathrm{2}} +{pz}+{q}\right)\left({z}^{\mathrm{2}} −{pz}+\frac{\mathrm{3}}{{q}}\right)=\mathrm{0} \\ $$$$\Rightarrow…
Question Number 142577 by loveineq last updated on 02/Jun/21 $$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{1}.\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{c}^{\mathrm{2}} +\mathrm{1}}\:\geqslant\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by ajfour last updated on…
Question Number 77041 by mathocean1 last updated on 02/Jan/20 $$\mathrm{solve}\:\mathrm{in}\:\left[\mathrm{0};\pi\right] \\ $$$$\mathrm{sin}{x}−\mathrm{sin}^{\mathrm{3}} {x}=\mathrm{1}−\mathrm{cos2}{x} \\ $$ Answered by mind is power last updated on 02/Jan/20 $$\Leftrightarrow…
Question Number 142573 by mnjuly1970 last updated on 02/Jun/21 Answered by qaz last updated on 02/Jun/21 $$\mathrm{csc}\:\mathrm{x} \\ $$ Commented by mnjuly1970 last updated on…
Question Number 77036 by Master last updated on 02/Jan/20 Commented by mr W last updated on 02/Jan/20 $${if}\:{f}\left(−{x}\right)=−{f}\left({x}\right), \\ $$$$\int_{−{a}} ^{+{a}} {f}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow\int_{−\mathrm{1}} ^{+\mathrm{1}}…
Question Number 11501 by @ANTARES_VY last updated on 27/Mar/17 $$\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{7}\boldsymbol{\mathrm{x}}−\mathrm{5}\right)\left(\mathrm{5}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{13}\boldsymbol{\mathrm{x}}+\mathrm{3}\right)\left(\mathrm{3}\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{8}\right)=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{all}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{multiples}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{real}}\:\:\boldsymbol{\mathrm{roots}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{equation}}. \\ $$ Commented by mrW1 last updated on…
Question Number 11500 by @ANTARES_VY last updated on 27/Mar/17 $$\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}−\mathrm{4}\right)\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{4}\right)=\mathrm{9} \\ $$$$\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{equation}}\:\:\boldsymbol{\mathrm{has}}\:\:\boldsymbol{\mathrm{multiple}}\:\:\boldsymbol{\mathrm{roots}}. \\ $$ Answered by ridwan balatif last updated on 27/Mar/17 $$\left(\left(\mathrm{x}^{\mathrm{2}}…
Question Number 77035 by necxxx last updated on 02/Jan/20 $${Solve}\:{for}\:{x}\:{in}: \\ $$$$\left({i}\right)\:\left(\mathrm{2}\left({x}+\mathrm{3}\right)−\mathrm{3}\left({x}−\mathrm{2}\right)\right)\left(\mathrm{2}{x}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\left({ii}\right)\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)\leqslant\mathrm{1} \\ $$ Answered by MJS last updated on 03/Jan/20 $$\left({i}\right) \\…
Question Number 142570 by mathsuji last updated on 02/Jun/21 $${Find}\:{all}\:{functions}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:{such}\:{that} \\ $$$${f}\left({x}+{y}\right)=\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}\left({y}\right)−\mathrm{4}{xyf}\left(\mathrm{2}{x}−\mathrm{3}{y}\right) \\ $$$$\left(\forall{x};{y}\in\mathbb{R}\right) \\ $$ Answered by ajfour last updated on 02/Jun/21 $${f}\left({x}\right)=\mathrm{5}{f}\left(\frac{{x}}{\mathrm{2}}\right)−{x}^{\mathrm{2}} {f}\left(−\frac{{x}}{\mathrm{2}}\right)…