Question Number 227393 by fantastic2 last updated on 20/Jan/26 $${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{…}}}} \\ $$$${x}=? \\ $$ Answered by Ghisom_ last updated on 20/Jan/26 $${x}^{\mathrm{2}} +{x}+\mathrm{1}={y} \\…
Question Number 227386 by fantastic2 last updated on 19/Jan/26 Answered by mr W last updated on 23/Jan/26 Commented by mr W last updated on 23/Jan/26…
Question Number 227361 by Hanuda354 last updated on 18/Jan/26 Commented by Hanuda354 last updated on 18/Jan/26 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{area},\:\mathrm{where} \\ $$$$\mathrm{A}\:=\:\mathrm{B}\:=\:\mathrm{24}.\mathrm{8}\:\mathrm{square}\:\mathrm{units}. \\ $$$$\left(\mathrm{A},\:\mathrm{B}\:\mathrm{are}\:\mathrm{shaded}\:\mathrm{regions}\right) \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{square}\:\mathrm{area}. \\ $$…
Question Number 227353 by fantastic2 last updated on 17/Jan/26 Answered by mr W last updated on 19/Jan/26 Commented by mr W last updated on 19/Jan/26…
Question Number 227341 by fantastic2 last updated on 17/Jan/26 $${find}\:{the}\:{surface}\:{area}\:{of}\:{a}\:{spherical} \\ $$$${cap} \\ $$ Commented by Kassista last updated on 17/Jan/26 $$ \\ $$$${are}\:{you}\:{simply}\:{asking}\:{to}\:{prove}\:{the}\:{surface}\:{area} \\…
Question Number 227329 by Lara2440 last updated on 17/Jan/26 $$\left.\mathrm{1}\right)\:\mathrm{Does}\:\mathrm{half}\:\mathrm{open}\:\mathrm{interval}\:{A}=\left[\mathrm{0},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{Compact}\:\mathrm{Space}\:? \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Prove}\:\mathrm{for}\:\mathrm{a}\:\:\mathrm{Compact}\:\mathrm{Space}\:{X}_{{k}} \:\: \\ $$$$\mathrm{Product}\:\mathrm{Space}\:{X}=\underset{{k}} {\prod}\:\left\{{X}_{{k}} ^{\:} \:;\:{k}\in{I}\right\}\:\:\mathrm{also}\:\mathrm{Compact}\:\mathrm{Space} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{a}\:\mathrm{Compact}\:\mathrm{Space}\:{X}\:\mathrm{and}\:\mathrm{Continuous}\:\mathrm{function}\:{f} \\ $$$$\mathrm{if}\:\:{f}\:\:\mathrm{satisfy}\:\:{f};{X}\rightarrow{Y}\:,\mathrm{Image}\:{Y}\:\mathrm{also}\:\mathrm{Compact}\:\mathrm{Space} \\ $$ Terms…
Question Number 227319 by Lara2440 last updated on 15/Jan/26 $${f}\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)={z} \\ $$$${f}\left({z}\right)=? \\ $$ Answered by breniam last updated on 14/Jan/26 $$\left(\forall{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\}\right)\left({f}\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}}…
Question Number 227274 by Lara2440 last updated on 11/Jan/26 $$\mathrm{Does}\:\mathrm{condition} \\ $$$$\underset{{x}\in\mathbb{R}} {\mathrm{sup}}\:\mid\mid{f}^{\left(\mathrm{1}\right)} \left({x}\right)\mid\mid<\infty\:,\:\:\mathrm{Gaurantee}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{Bounded}\:\mathrm{in}\:\mathbb{R}? \\ $$ Answered by peace2 last updated on 12/Jan/26 $${non}\:{f}\left({x}\right)={x} \\…
Question Number 227283 by Edwanrosario201444 last updated on 11/Jan/26 $${Proof}\:{that}\:\sqrt{\mathrm{2}}\:{is}\:{irrational} \\ $$ Answered by TonyCWX last updated on 11/Jan/26 $$\mathrm{Suppose}\:\sqrt{\mathrm{2}}\:\mathrm{is}\:\mathrm{rational} \\ $$$$\mathrm{Let}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\sqrt{\mathrm{2}}\:=\:\frac{\mathrm{a}}{\mathrm{b}} \\…
Question Number 227236 by AlanMuhamad last updated on 09/Jan/26 $$\mid{x}−\mathrm{1}\mid\geqslant\mathrm{6} \\ $$$${x}−\mathrm{1}\geqslant\mathrm{6}\:\:\:\:\:{or}\:\:\:\:\:{x}−\mathrm{1}\leqslant−\mathrm{6} \\ $$$${x}\geqslant\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\leqslant−\mathrm{5} \\ $$$${s}_{\mathrm{1}} =\left[\mathrm{7},+\infty\right)\:\:\:\:\:\:\:\:{s}_{\mathrm{2}} =\left(−\infty,−\mathrm{5}\right] \\ $$$${s}=\left(−\infty,−\mathrm{5}\right]\:{U}\:\left[\mathrm{7},+\infty\right) \\ $$$${R}/\left(−\mathrm{5}\:,\:\mathrm{7}\:\right) \\ $$ Terms…