Question Number 222097 by MathematicalUser2357 last updated on 17/Jun/25 Commented by mr W last updated on 17/Jun/25 $${wrong}! \\ $$$$\left({a}+{b}+{c}\right)^{{n}} =\underset{\underset{\mathrm{0}\leqslant{i},{j},{k}\leqslant{n}} {{i}+{j}+{k}={n}}} {\sum}\left(\frac{{n}!}{{i}!{j}!{k}!}{a}^{{i}} {b}^{{j}} {c}^{{k}}…
Question Number 221787 by efronzo1 last updated on 10/Jun/25 $$\:\:\sqrt{\frac{\mathrm{1}−\mathrm{4x}\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}{\mathrm{2}}}\:=\:\mathrm{1}−\mathrm{8x}^{\mathrm{2}} \\ $$$$\:\:\mathrm{x}=?\: \\ $$ Answered by mr W last updated on 11/Jun/25 $$\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} =\mathrm{1}−\left(\mathrm{2}{x}\right)^{\mathrm{2}}…
Question Number 221247 by fantastic last updated on 28/May/25 Commented by fantastic last updated on 28/May/25 $${what}\:{is}\:{the}\:{yellow}\:{chord}\:{length}? \\ $$ Answered by mehdee7396 last updated on…
Question Number 220737 by Spillover last updated on 18/May/25 Answered by som(math1967) last updated on 18/May/25 $$\:\mathrm{0}.\overset{.} {\mathrm{6}}=\mathrm{0}.\mathrm{6}+\mathrm{0}.\mathrm{06}+\mathrm{0}.\mathrm{006}+… \\ $$$$=\frac{\mathrm{0}.\mathrm{6}}{\mathrm{1}−.\mathrm{1}}=\frac{\mathrm{6}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Commented by Spillover…
Question Number 220738 by Spillover last updated on 18/May/25 Answered by mr W last updated on 18/May/25 $${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}} +…=\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}}={k} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{2}}…
Question Number 220159 by Nicholas666 last updated on 06/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x},\:{y}\:\in\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}\:+\:\left({x}\:+\:{y}\right)^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:{y}^{\mathrm{3}} }}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\leqslant\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} }}\:+\:\frac{\mathrm{2}}{\:^{\mathrm{4}} \sqrt{\mathrm{1}\:+\:{x}^{\mathrm{5}}…
Question Number 220160 by Nicholas666 last updated on 06/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x}\:,\:{y}\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\left[\:\frac{\sqrt[{\mathrm{3}\:\:}]{\boldsymbol{{x}}^{\mathrm{3}} \:+\:\boldsymbol{{y}}^{\mathrm{3}} \:+\:\boldsymbol{\zeta}\left(\mathrm{3}\right)}}{\mathrm{1}\:+\:\boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} } \:}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{x}}^{\mathrm{4}} +\:\boldsymbol{\Gamma}\left(\boldsymbol{{y}}+\mathrm{1}\right)}}{\left(\mathrm{1}\:+\:\boldsymbol{{y}}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} }\:+\:\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{5}} \:+\:\boldsymbol{{y}}^{\mathrm{5}} \right)}{\:\sqrt{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}}…
Question Number 220096 by Nicholas666 last updated on 05/May/25 $$ \\ $$$$\:\:\:\mathrm{let}\:{a},\:{b},\:{c},\:{d},\:{e}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{K}\:=\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:+\:{e}\:+\mathrm{1}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\underset{\boldsymbol{{cyc}}} {\sum}\:\frac{\mathrm{1}}{\boldsymbol{{k}}−\boldsymbol{{a}}}\:<\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(\frac{\sqrt[{\mathrm{4}\:\:\:\:}]{\boldsymbol{{e}}^{\mathrm{3}} \boldsymbol{{d}}^{\mathrm{3}} \boldsymbol{{c}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{a}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:\:}]{\boldsymbol{{d}}^{\:\mathrm{3}} \boldsymbol{{c}}^{\mathrm{2}}…
Question Number 219874 by mr W last updated on 03/May/25 $${find}\:\sqrt{\mathrm{2}^{\mathrm{6}^{\mathrm{2}^{\mathrm{1}^{\mathrm{4}^{\mathrm{4}} } } } } }=? \\ $$ Answered by fantastic last updated on 03/May/25…
Question Number 219933 by BaliramKumar last updated on 03/May/25 $$\mathrm{A}^{\mathrm{1}} \:+\:\mathrm{B}^{\mathrm{2}} \:+\:\mathrm{C}^{\mathrm{3}} \:+\:\mathrm{D}^{\mathrm{4}} \:=\:\overline {\mathrm{ABCD}} \\ $$$${find}\:\:{ABCD} \\ $$ Answered by MrGaster last updated on…