Question Number 220468 by Rojarani last updated on 13/May/25 $$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\left({a},{b},{c}\:\in{R}\right) \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$ Answered by…
Question Number 220469 by Nicholas666 last updated on 13/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{6}{x}\left(\mathrm{1}\:−\:{x}\right)}{\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\mathrm{ln}\:\left({x}\:+\:\mathrm{1}\right)}\:{dx} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 220499 by Jubr last updated on 13/May/25 Commented by Jubr last updated on 13/May/25 $${Find}\:{x}\:{and}\:{y} \\ $$ Answered by Ghisom last updated on…
Question Number 220493 by leromain last updated on 13/May/25 $${Let}\:{ABC}\:{be}\:{a}\:{triangle}\:{such} \\ $$$${CosA}+{cosB}+{cosC}=\sqrt{\mathrm{2}} \\ $$$${SinA}+{sinB}+{sinC}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$${Find}\:{A},{B},{C} \\ $$ Commented by mr W last updated on…
Question Number 220495 by Jubr last updated on 13/May/25 Commented by mr W last updated on 14/May/25 $${i}\:{guess}\:{even}\:{you}\:{also}\:{don}'{t}\:{know} \\ $$$${what}\:{the}\:{question}\:{means}\:{with}\:{all} \\ $$$${the}\:{arrows}.\:{when}\:{the}\:{question}\:{is} \\ $$$${unclear},\:{it}\:{can}\:{not}\:{be}\:{solved}.\:{so} \\…
Question Number 220486 by hardmath last updated on 13/May/25 $$\mathrm{z}\:\in\:\mathbb{C}\:\:\:\mathrm{and}\:\:\:\lambda\:>\:\mathrm{0} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{z}\:+\:\mathrm{2}\lambda\mid\:+\:\mid\mathrm{z}\:+\:\lambda\mid\:\geqslant\:\mid\mathrm{z}\:+\:\frac{\mathrm{3}\lambda\:−\:\lambda\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{2}}\mid \\ $$ Commented by MrGaster last updated on 14/May/25 The original problem is equivalent to: In a plane, the sum of the distances from any point to two vertices of an equilateral triangle is greater than the distance from that point to the third vertex. This can be easily proven. Then Ptolemy's theorem can be applied. Commented…
Question Number 220480 by SdC355 last updated on 13/May/25 $$\mathrm{Can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{teach}\:\mathrm{me}\:\mathrm{about} \\ $$$$\mathrm{Weber}\:\mathrm{function}\:\boldsymbol{\mathrm{E}}_{\nu} \left({z}\right)\:\mathrm{and}\:\mathrm{Anger}\:\mathrm{function}\:\boldsymbol{\mathrm{J}}_{\nu} \left({z}\right)?? \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{Consider}\:{n}-\mathrm{dimensional}\:\mathrm{Euclidean}\:\mathrm{Space} \\ $$$$\mathrm{and}\:\mathrm{function}\:{f}\:,\:{f};\mathbb{R}^{{n}} \rightarrow\mathbb{R} \\ $$$$\mathrm{Helmholt}{z}\:\mathrm{Equation}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\left(\bigtriangledown^{\mathrm{2}}…
Question Number 220405 by MathematicalUser2357 last updated on 12/May/25 $$\mathrm{9}^{{x}^{\mathrm{2}} } −\mathrm{3}^{{x}+\mathrm{1}} =\mathrm{0} \\ $$ Answered by MathematicalUser2357 last updated on 12/May/25 $${x}=\mathrm{1} \\ $$…
Question Number 220403 by MrGaster last updated on 12/May/25 Answered by MrGaster last updated on 12/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 220396 by MrGaster last updated on 12/May/25 Commented by MathematicalUser2357 last updated on 12/May/25 The second one has to be an approximation. Commented by hardmath last updated on 12/May/25 $$\mathrm{nice}\:\mathrm{problem}\:\mathrm{dear}\:\mathrm{professor}…