Question Number 1744 by Rasheed Ahmad last updated on 13/Sep/15 $${If}\:\boldsymbol{\mathrm{A}}\:{and}\:\boldsymbol{\mathrm{B}}\:{are}\:{two}\:{sets}\:{and}\:\mathbb{U}\:{is} \\ $$$${a}\:{universal}\:{set}\:{prove}\:{that} \\ $$$$\boldsymbol{\mathrm{A}}\:\subseteq\:\boldsymbol{\mathrm{B}}\:\:\Rightarrow\:\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\:\cup\:\left(\boldsymbol{\mathrm{A}}'\:\cap\:\boldsymbol{\mathrm{B}}\right) \\ $$ Answered by Rasheed Ahmad last updated on 19/Sep/15…
Question Number 1743 by 123456 last updated on 12/Sep/15 $${u}'\left({v}−{v}'\right)+{uv}'=\mathrm{0} \\ $$$${u}=? \\ $$ Commented by Rasheed Ahmad last updated on 13/Sep/15 $${u}'\left({v}−{v}'\right)+{uv}'=\mathrm{0} \\ $$$${u}'{v}+{uv}'−{u}'{v}'=\mathrm{0}…
Question Number 1740 by tabrez8590@gmail last updated on 11/Sep/15 $${how}\:\:\mathrm{0}!=\mathrm{1} \\ $$ Commented by 123456 last updated on 11/Sep/15 $$\mathrm{1}!=\mathrm{1}\centerdot\mathrm{0}! \\ $$$$\mathrm{so}\:\mathrm{for}\:\mathrm{it}\:\mathrm{work}\:\mathrm{to}\:\mathrm{1}\:\mathrm{them}\:\mathrm{0}!=\mathrm{1} \\ $$ Answered…
Question Number 132809 by mr W last updated on 17/Feb/21 Commented by mr W last updated on 18/Feb/21 $${a}\:{mountain}\:{has}\:{the}\:{shape}\:{of}\:{a}\:{right} \\ $$$${circular}\:{cone}\:{as}\:{shown}\:\left({h}>\sqrt{\mathrm{3}}{r}\right). \\ $$$${from}\:{point}\:{A}\:{to}\:{point}\:{B}\:{two} \\ $$$${sightseeing}\:{roads}\:{one}\:{time}\:{around}…
Question Number 1738 by Rasheed Sindhi last updated on 08/Sep/15 $${Let}\:\boldsymbol{\mathrm{a}}\:{belongs}\:{to}\:{an}\:{interval}\:\boldsymbol{\mathrm{A}},\boldsymbol{\mathrm{k}}\:{is}\:{aconstant}\:{such}\:{that}\: \\ $$$$\boldsymbol{\mathrm{k}}\in\mathbb{R}^{+} \:{and}\:\boldsymbol{\mathrm{k}}<\boldsymbol{\mathrm{a}}\:. \\ $$$${Find}\:{out}\:\boldsymbol{\mathrm{A}}\:{in}\:{case}: \\ $$$$\:\:\:\:\:\:\:\:\left(\:\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{k}}\right)^{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{k}}} \geqslant\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{k}}\right)^{\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{k}}} \\ $$ Terms of Service Privacy…
Question Number 1736 by navajyoti.tamuli.tamuli@gmail. last updated on 04/Sep/15 $${Express}\:\mathrm{0}.\mathrm{999}…\:{as}\:\frac{{p}}{{q}} \\ $$ Answered by 123456 last updated on 04/Sep/15 $${x}=\mathrm{0}.\mathrm{999999}…. \\ $$$$\mathrm{10}{x}=\mathrm{9}.\mathrm{9999999}….. \\ $$$$\mathrm{9}{x}=\mathrm{9} \\…
Question Number 1734 by Rasheed Ahmad last updated on 07/Sep/15 $$\boldsymbol{\mathrm{a}}\:\:{belongs}\:{to}\:{an}\:{interval}\:\boldsymbol{\mathrm{A}}\:{and} \\ $$$$\boldsymbol{\mathrm{k}}\in\mathbb{R}^{+} \:\:{is}\:{a}\:{constant}.\:{Determine} \\ $$$$\boldsymbol{\mathrm{A}}\:{in}\:{case} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{k}}\mid^{\mid\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{k}}\:\mid} \geqslant\:\mid\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{k}}\mid^{\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{k}}\mid} \: \\ $$$$ \\ $$ Terms…
Question Number 1729 by Rasheed Ahmad last updated on 04/Sep/15 $${Prove}/{disprove}/{prove}\:{for}\:{an} \\ $$$${interval}\:{as}\:{the}\:{case}\:{may}\:{be}: \\ $$$$\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:\overset{?} {<}\:\left\{\left({x}+\mathrm{1}\right)!\right\}^{\frac{\mathrm{1}}{{x}+\mathrm{1}}} \:\:,\:{x}\in\mathbb{N}\:\left[{x}\neq\mathrm{0}\right] \\ $$$$\left({Generalization}\:{of}\:{Q}\:\mathrm{1700}\right) \\ $$ Commented by 123456…
Question Number 132802 by Algoritm last updated on 16/Feb/21 Commented by MJS_new last updated on 16/Feb/21 $$\mathrm{you}\:\mathrm{must}\:\mathrm{approximate} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 132799 by Ñï= last updated on 16/Feb/21 $$\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}+{ix}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}−{ix}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \int\frac{{d}\left({e}^{{i}\frac{\pi}{\mathrm{4}}} {x}\right)}{\mathrm{1}+\left({e}^{{i}\frac{\pi}{\mathrm{4}}} {x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \int\frac{{d}\left({e}^{−{i}\frac{\pi}{\mathrm{4}}} \right){x}}{\mathrm{1}+\left({e}^{−{i}\frac{\pi}{\mathrm{4}}} {x}\right)^{\mathrm{2}} }…