Question Number 24605 by ajfour last updated on 22/Nov/17 Commented by ajfour last updated on 22/Nov/17 $${Find}\:{the}\:{central}\:{octagonal}\:{area} \\ $$$${if}\:{ABCD}\:{is}\:{a}\:{square}\:{and}\:{points} \\ $$$${P},\:{Q},\:{R},\:{and}\:{S}\:{are}\:{the}\:{midpoints} \\ $$$${of}\:{its}\:{sides}.\:{Take}\:{edge}\:{length}\:{of} \\ $$$${square}\:{to}\:{be}\:{unity}.…
Question Number 90122 by M±th+et£s last updated on 21/Apr/20 $${each}\:{vertex}\:{of}\:{a}\:{cube}\:{is}\:{to}\:{be}\:{labeled} \\ $$$${with}\:{an}\:{integer}\:\mathrm{1}\:{through}\:\mathrm{8}\:,\:{with} \\ $$$${each}\:{integer}\:{being}\:{used}\:{once},{in}\:{such} \\ $$$${a}\:{way}\:{that}\:{the}\:{sum}\:{of}\:{the}\:{four}\:{numbers} \\ $$$${on}\:{the}\:{vertices}\:{of}\:{a}\:{face}\:{is}\:{the}\:{same}\: \\ $$$${for}\:{each}\:{face}.{Arrangements}\:{that}\:{can} \\ $$$${be}\:{obtained}\:{from}\:{each}\:{other}\:{through} \\ $$$${rotations}\:{of}\:{the}\:{cube}\:{are}\:{considered} \\…
Question Number 155625 by puissant last updated on 02/Oct/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{tanx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\:? \\ $$ Answered by yeti123 last updated on 03/Oct/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{tan}\:{x}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}…
Question Number 90087 by john santu last updated on 21/Apr/20 $$\underset{{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{{k}} }\:=\:? \\ $$ Commented by frc2crc last updated on 22/Apr/20 $${its}\:{around}\:\mathrm{1}.\mathrm{29}… \\…
Question Number 155574 by peter frank last updated on 02/Oct/21 $$\mathrm{The}\:\mathrm{mean}\:\mathrm{and}\:\mathrm{standard}\:\mathrm{deviation} \\ $$$$\mathrm{of}\:\mathrm{20}\:\mathrm{observation}\:\:\mathrm{are}\:\mathrm{found}\:\mathrm{to}\:\mathrm{be}\: \\ $$$$\mathrm{10}\:\mathrm{and}\:\mathrm{2}\:\mathrm{respectively}\:.\mathrm{On}\:\mathrm{rechecking} \\ $$$$\mathrm{it}\:\mathrm{was}\:\mathrm{found}\:\mathrm{that}\:\:\mathrm{an}\:\mathrm{observation} \\ $$$$\mathrm{8}\:\mathrm{was}\:\mathrm{incorrect}.\mathrm{Calculate}\:\mathrm{the}\:\mathrm{incorrect} \\ $$$$\mathrm{mean}\:\mathrm{and}\:\mathrm{standard}\:\mathrm{deviation} \\ $$$$\left(\boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{I}}\mathrm{f}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{wrong}}\:\boldsymbol{\mathrm{iterm}}\:\boldsymbol{\mathrm{was}}\:\:\boldsymbol{\mathrm{ommited}} \\ $$$$\left(\boldsymbol{\mathrm{b}}\right)\:\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{replaced}}\:\boldsymbol{\mathrm{by}}\:\mathrm{12}…
Question Number 24111 by ajfour last updated on 12/Nov/17 Commented by ajfour last updated on 12/Nov/17 $${Solution}\:{to}\:{Q}.\:\mathrm{24100} \\ $$ Answered by ajfour last updated on…
Question Number 89560 by peter frank last updated on 18/Apr/20 $${without}\:{use}\:{intergration} \\ $$$${by}\:{party} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {e}^{\theta} \mathrm{cos}\:\mathrm{2}\theta\:{d}\theta \\ $$ Commented by mathmax by abdo…
Question Number 89493 by M±th+et£s last updated on 17/Apr/20 Commented by M±th+et£s last updated on 17/Apr/20 $${what}\:{is}\:{the}\:{solution}\:{if}\:{the}\:{cable}\:{was} \\ $$$$\mathrm{200}\:{m}\:{long}? \\ $$ Terms of Service Privacy…
Question Number 23564 by NECx last updated on 01/Nov/17 $${The}\:{curved}\:{surface}\:{area}\:{of}\:{a}\:{cone} \\ $$$${is}\:\mathrm{21}{cm}^{\mathrm{2}} \:.{Calculate}\:{the}\:{curved} \\ $$$${surface}\:{area}\:{of}\:{a}\:{similar}\:{cone} \\ $$$${whose}\:{height}\:{is}\:\mathrm{4}\:{times}\:{the}\:{other}. \\ $$ Commented by NECx last updated on…
Question Number 88997 by jagoll last updated on 14/Apr/20 Commented by john santu last updated on 14/Apr/20 $$\mathrm{4}{R}+\mathrm{2}{R}\sqrt{\mathrm{2}}\:=\:\mathrm{8}\sqrt{\mathrm{2}} \\ $$$${R}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\:=\:\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${R}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:×\:\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\: \\ $$$${R}=\:\frac{\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{8}}{\mathrm{4}−\mathrm{2}}\:=\:\mathrm{4}\sqrt{\mathrm{2}}\:−\mathrm{4}\:{dm} \\…