Question Number 168073 by pete last updated on 02/Apr/22 $$\mathrm{The}\:\mathrm{unit}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{a}\:\mathrm{two}\:\mathrm{degit}\:\mathrm{number}\:\mathrm{is}\:\mathrm{1}\:\mathrm{less} \\ $$$$\mathrm{than}\:\mathrm{the}\:\mathrm{tens}\:\mathrm{degit}.\:\mathrm{if}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{increased} \\ $$$$\mathrm{by}\:\mathrm{8}\:\mathrm{and}\:\mathrm{then}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{digits},\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{8}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{number}. \\ $$ Answered by som(math1967) last updated on 02/Apr/22…
Question Number 168054 by Mastermind last updated on 01/Apr/22 $${Simplify}\:\:\frac{\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}+\sqrt{{x}}} \\ $$ Commented by MJS_new last updated on 01/Apr/22 $$\mathrm{what}\:\mathrm{do}\:\mathrm{you}\:\mathrm{want}?\:\mathrm{what}\:\mathrm{is}\:“\mathrm{simple}''? \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{do}\:\mathrm{this}: \\ $$$$\frac{\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}+\sqrt{{x}}}=\frac{\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}+\sqrt{{x}}}×\frac{\sqrt{\mathrm{1}−{x}}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}−\sqrt{{x}}}= \\…
Question Number 168043 by Sigunur last updated on 01/Apr/22 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\frac{\mathrm{7x}^{\mathrm{2}} \:+\:\mathrm{7}}{\:\sqrt{\mathrm{7x}^{\mathrm{2}} \:+\:\mathrm{3}}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{18}}{\:\sqrt{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{17}}}\:+\:\frac{\mathrm{3z}^{\mathrm{2}} \:+\:\mathrm{26}}{\:\sqrt{\mathrm{3z}^{\mathrm{2}} \:+\:\mathrm{1}}}\:=\:\mathrm{16} \\ $$ Commented by MJS_new last updated…
Question Number 168042 by cortano1 last updated on 01/Apr/22 $$\:\:{The}\:{number}\:{of}\:{natural}\:{number}\:{k} \\ $$$$\:{such}\:{that}\:{k}\mid\:{n}^{\mathrm{7}} −{n}\:{for}\:\forall{n}\:{natural} \\ $$$$\:{numbers} \\ $$ Commented by Rasheed.Sindhi last updated on 02/Apr/22 $$\mathcal{D}{epends}\:{upon}\:{individual}\:{n}.…
Question Number 102490 by ajfour last updated on 09/Jul/20 $${x}^{\mathrm{3}} −{bx}−{c}=\mathrm{0}\:\:\:\:\:;\:\:{b},\:{c}\:>\mathrm{0}\:;\:\:\left(\frac{{b}}{\mathrm{3}}\right)^{\mathrm{3}} >\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${To}\:{find}\:{the}\:{three}\:{real}\:{roots}\:{without} \\ $$$${the}\:{use}\:{of}\:{trigonometric}\:{solution} \\ $$$${to}\:{cubic}\:{polynomial}… \\ $$ Answered by ajfour last updated…
Question Number 36926 by maxmathsup by imad last updated on 07/Jun/18 $${let}\:{f}\left({x}\right)\:=\:{e}^{−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}^{\left({n}\right)} \left({x}\right)={p}_{{n}} \left({x}\right){e}^{−{x}^{\mathrm{2}} } \:\:{with}\:{p}_{{n}} \:{is}\:{a}\:{polynom} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{relation}\:{of}\:{recurrence}\:{between}\:{the}\:{p}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{p}_{\mathrm{1}} ,{p}_{\mathrm{2}}…
Question Number 167994 by mathlove last updated on 31/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167985 by peter frank last updated on 31/Mar/22 $$\mathrm{If}\:\:\mathrm{af}\left(\mathrm{x}\right)+\mathrm{bf}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{where}\: \\ $$$$\mathrm{a}\neq\mathrm{b}\:\mathrm{and}\:\:\mathrm{x}\neq\mathrm{0}\:\mathrm{show}\:\mathrm{that} \\ $$$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\right)\left(\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{x}}}−\boldsymbol{\mathrm{bx}}\right) \\ $$ Answered by som(math1967) last updated on…
Question Number 36911 by prof Abdo imad last updated on 07/Jun/18 $${p}\:{is}\:{a}\:{polynome}\:{having}\:{nroots}\:{simples} \\ $$$${x}_{{i}} \:\left(\mathrm{1}\leqslant{x}_{{i}} \leqslant{n}\:\right)\:{with}\:{x}_{{i}} ^{\mathrm{2}} \:\neq\mathrm{1}\:\:{calculste} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:. \\ $$…
Question Number 36909 by prof Abdo imad last updated on 07/Jun/18 $${let}\:{p}\left({x}\right)={x}^{\mathrm{3}} \:−\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}\:{and}\:\alpha\:{is}\:{root}\:{of}\:{p}\left({x}\right) \\ $$$${prove}\:{that}\:\alpha\notin\:{Q}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com