Question Number 122666 by mathocean1 last updated on 18/Nov/20 $${we}\:{suppose}\:{that}\:{today}\:{is}\: \\ $$$${Monday}\:\mathrm{05}^{{th}} {November}\:\mathrm{2020}. \\ $$$$\mathrm{2020}\:{is}\:{a}\:{bissextile}\:{year}\:\left({it}\:\right. \\ $$$$\left.{contains}\:\mathrm{366}\:{days}\right).\:{John}\:{was} \\ $$$${born}\:{the}\:\mathrm{11}^{{th}} {mars}\:\mathrm{1972}. \\ $$$${find}\:{the}\:{day}\:{of}\:{week}\:{he}\:{was} \\ $$$${born}. \\…
Question Number 122659 by mathocean1 last updated on 18/Nov/20 $${Determinate}\:{the}\:{geometric} \\ $$$${aspect}\:{described}\:{by}\:{M}\left({z}\right)\:{in} \\ $$$${complex}\:{plane}\:{such}\:{that}: \\ $$$${arg}\left(\overset{−} {{z}}−\mathrm{3}+{i}\right)\equiv\frac{\pi}{\mathrm{4}}\left[\mathrm{2}\pi\right] \\ $$ Answered by MJS_new last updated on…
Question Number 57051 by Hassen_Timol last updated on 29/Mar/19 $$ \\ $$$$\:\:\boldsymbol{{F}}_{\boldsymbol{{n}}} \:=\:\frac{\varphi^{{n}} \:−\:\left(\mathrm{1}\:−\:\varphi\right)^{{n}} }{\:\sqrt{\mathrm{5}}}\:,\:\:\mathrm{with}\:\begin{cases}{\varphi\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$$$\:\:\left(\boldsymbol{{G}}_{\boldsymbol{{n}}} \right)\::\:\begin{cases}{\boldsymbol{{G}}_{\mathrm{1}} \:=\:\mathrm{1}}\\{\boldsymbol{{G}}_{\mathrm{2}} \:=\:\mathrm{1}}\\{\boldsymbol{{G}}_{{m}} \:=\:\boldsymbol{{G}}_{\mathrm{m}−\mathrm{1}} \:+\:\boldsymbol{{G}}_{{m}−\mathrm{2}} }\end{cases}…
Question Number 188098 by MathsFan last updated on 25/Feb/23 $$\int\boldsymbol{\mathrm{x}}!\boldsymbol{\mathrm{dx}} \\ $$ Answered by aba last updated on 25/Feb/23 $$\int{x}!{dx}=\int\Gamma\left({x}+\mathrm{1}\right){dx}=\int\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\mathrm{x}−\mathrm{1}} \mathrm{e}^{−{t}} {dtdx} \\…
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Question Number 57023 by otchereabdullai@gmail.com last updated on 28/Mar/19 $$\mathrm{There}\:\mathrm{are}\:\mathrm{28}\:\mathrm{players}\:\mathrm{in}\:\mathrm{a}\:\mathrm{national}\: \\ $$$$\mathrm{football}\:\mathrm{team}.\:\mathrm{14}\:\mathrm{play}\:\mathrm{midfield}\:\mathrm{and} \\ $$$$\mathrm{defence},\:\mathrm{15}\:\mathrm{play}\:\mathrm{defence}\:\mathrm{and}\:\mathrm{attack}\: \\ $$$$\mathrm{and}\:\mathrm{3}\:\mathrm{play}\:\mathrm{midfield}\:\mathrm{only}.\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{players}\:\mathrm{who}\:\mathrm{play}\:\mathrm{attack}\:\mathrm{only}\:\mathrm{is}\: \\ $$$$\mathrm{twice}\:\mathrm{those}\:\mathrm{who}\:\mathrm{play}\:\mathrm{defence}\:\mathrm{only},\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{who}\:\mathrm{play}\:\mathrm{defence} \\ $$$$\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{those}\:\mathrm{who}\:\mathrm{play}\:\mathrm{attack}.\:\mathrm{If}\: \\…
Question Number 122555 by ZiYangLee last updated on 18/Nov/20 $${prove}\:{that}\begin{vmatrix}{{a}}&{\mathrm{1}}&{{a}^{\mathrm{2}} \left({b}+{c}\right)}\\{{b}}&{\mathrm{1}}&{{b}^{\mathrm{2}} \left({c}+{a}\right)}\\{{c}}&{\mathrm{1}}&{{c}^{\mathrm{2}} \left({a}+{b}\right)}\end{vmatrix}=\mathrm{0} \\ $$ Commented by bramlexs22 last updated on 18/Nov/20 $$\Rightarrow{a}\:\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:{b}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:\:\:\:{ac}^{\mathrm{2}}…
Question Number 57007 by naka3546 last updated on 28/Mar/19 $$\mathrm{cosec}\:\left(\frac{\pi}{\mathrm{14}}\right)\:−\:\mathrm{4}\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:\:=\:\:? \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 29/Mar/19 Commented by tanmay.chaudhury50@gmail.com last updated on…
Question Number 122540 by help last updated on 18/Nov/20 Commented by help last updated on 18/Nov/20 $${find}\:{the}\:{perimeter}\:{of}\:{the}\:{shape} \\ $$ Answered by Olaf last updated on…
Question Number 188079 by 073 last updated on 25/Feb/23 Commented by 073 last updated on 25/Feb/23 $$\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\right)=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=? \\ $$$$\mathrm{please}\:\mathrm{solution}…