Question Number 122679 by mathocean1 last updated on 18/Nov/20 $${N}=\mathrm{1335}\:{in}\:{base}\:{a}.\: \\ $$$${Write}\:{N}\:{in}\:{base}\:{a}+\mathrm{1} \\ $$ Answered by mr W last updated on 18/Nov/20 $${N}=\left(\mathrm{1335}\right)_{{a}} \\ $$$$=\mathrm{1}×{a}^{\mathrm{3}}…
Question Number 122673 by mathocean1 last updated on 18/Nov/20 $${Peter}\:{invite}\:{his}\:{son}\:{to}\:\:{choose} \\ $$$${secret}\:{integer}\:{n}\:{such}\:{that}\: \\ $$$${n}\:\in\:\left[\mathrm{1};\mathrm{63}\right]\:{and}\:{to}\:{choose}\:{among} \\ $$$${these}\:\mathrm{6}\:{following}\:{card}\:\left({CARTE}\right) \\ $$$${the}\:{card}\:{which}\:{contains}\:{the} \\ $$$${choosed}\:{secret}\:{digit}.\:{i}\:{will} \\ $$$${link}\:{these}\:{cards}\:{in}\:{image}. \\ $$$$\left.\mathrm{1}\right)\:{How}\:{should}\:{the}\:{Peter}'{s}\:{son} \\…
Question Number 122666 by mathocean1 last updated on 18/Nov/20 $${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…