Question Number 204621 by hardmath last updated on 23/Feb/24 $$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}\:\in\:\mathbb{R}^{+} \\ $$$$\mathrm{If}\:\:\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$ Answered by A5T last updated on 23/Feb/24 $$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\left(\frac{\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}}\Rightarrow{a}+{b}+{c}\geqslant\frac{\mathrm{1}}{\mathrm{3}}…
Question Number 204610 by mnjuly1970 last updated on 23/Feb/24 $$ \\ $$$$\:\:\:{If}\:,\:\:\:{f}\left({x}\right)\:=\:\begin{cases}{\:\mathrm{2}^{\mathrm{2}{x}} −\:{log}_{\mathrm{3}} \:\left(\:{x}+\mathrm{3}\:\right)\:\:\:\:;\:\:\:{x}\:\geqslant\mathrm{5}}\\{\:{f}\:\left(\mathrm{1}+\:{x}\:\right)\:\:−\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:;\:\:{x}\:<\:\mathrm{5}}\end{cases}\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\:{f}\:\left(\mathrm{0}\:\right)=\:? \\ $$$$ \\ $$ Answered by Rasheed.Sindhi…
Question Number 204590 by sphelele last updated on 22/Feb/24 Answered by A5T last updated on 22/Feb/24 $$\left.{a}\right)=\frac{\mathrm{2}^{{n}+\mathrm{2}+\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}^{\mathrm{3}{n}−\mathrm{3}} }=\frac{\mathrm{2}^{\mathrm{3}{n}+\mathrm{4}} }{\mathrm{2}^{\mathrm{3}{n}+\mathrm{4}} \mathrm{2}^{−\mathrm{7}} }=\mathrm{128} \\ $$$$\left.{b}\right)=\left(\frac{{b}−{a}}{{ab}}\overset{−\mathrm{1}} {\right)}=\frac{{ab}}{{b}−{a}}…
Question Number 204583 by hardmath last updated on 22/Feb/24 $$\mathrm{For}\:\:\:\mathrm{z}\:=\:\mathrm{a}\:−\:\mathrm{bi} \\ $$$$\mathrm{If}\:\:\:\left(\mid\mathrm{z}\mid\:−\:\mathrm{z}\right)\centerdot\left(\mid\mathrm{z}\mid\:+\:\overline {\mathrm{z}}\right)\:=\:\mathrm{4bi} \\ $$$$\mathrm{Find}\:\:\:\mid\mathrm{z}\mid\:=\:? \\ $$ Answered by A5T last updated on 22/Feb/24 $$\left(\mid{z}\mid−{z}\right)\left(\mid{z}\mid+\overset{−}…
Question Number 204545 by hardmath last updated on 21/Feb/24 $$\mathrm{If} \\ $$$$\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:\:+\:\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} } \\ $$$$\mathrm{b}\:=\:\mathrm{0},\mathrm{99} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{a}\:<\:\mathrm{b} \\ $$ Answered by…
Question Number 204544 by hardmath last updated on 21/Feb/24 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }\:\:+\:\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\:\:<\:\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$ Answered by mr W last updated on 21/Feb/24…
Question Number 204511 by Ghisom last updated on 20/Feb/24 $$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\mathrm{2}} −\mathrm{10}\lfloor{x}\rfloor+\frac{\mathrm{57}}{\mathrm{4}}=\mathrm{0} \\ $$ Answered by mr W last updated on 20/Feb/24 $$\lfloor{x}\rfloor={n}\:\in\:\mathbb{Z}^{+} \\…
Question Number 204510 by Ghisom last updated on 20/Feb/24 $$\mathrm{solve}\:\mathrm{for}\:{x}\neq{y}\wedge{y}\neq{z}\wedge{z}\neq{x} \\ $$$$\left(\mathrm{exact}\:\mathrm{solutions}\:\mathrm{required}\right) \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){x}}={y} \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){y}}={z} \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){z}}={x} \\ $$ Answered by Frix last updated…
Question Number 204512 by Ghisom last updated on 20/Feb/24 $$\mathrm{solve}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{3}^{\mathrm{2i}{x}} −\mathrm{3}^{\mathrm{i}{x}} \mathrm{2}+\mathrm{5}=\mathrm{0} \\ $$ Answered by Rasheed.Sindhi last updated on 20/Feb/24 $$\left(\mathrm{3}^{{ix}} \right)^{\mathrm{2}}…
Question Number 204500 by DeArtist last updated on 19/Feb/24 $$\mathrm{Given}\:\mathrm{that}\:{I}\:=\:\int\int_{{R}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} {dxdy}\:\mathrm{where}\:{R} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{region}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\:{a}^{\mathrm{2}} \\ $$$$\mathrm{use}\:\mathrm{a}\:\mathrm{suitable}\:\mathrm{transformation}\:\mathrm{to}\:\mathrm{evaluate}\:{I} \\ $$ Answered by witcher3…