Question Number 202535 by Rasheed.Sindhi last updated on 29/Dec/23 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}}\:=\mathrm{2} \\ $$$$\left(\mathrm{An}\:\mathrm{alteration}\:\mathrm{of}\:\mathrm{Q}#\mathrm{202500}\right) \\ $$ Answered by mr W last updated on…
Question Number 202562 by cherokeesay last updated on 29/Dec/23 Answered by ajfour last updated on 29/Dec/23 Commented by ajfour last updated on 29/Dec/23 Happy New Year! Commented…
Question Number 202511 by MrGHK last updated on 28/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202504 by MASANJAJJ last updated on 28/Dec/23 $$ \\ $$$$\mathrm{Tony}\:\mathrm{is}\:\mathrm{selling}\:\mathrm{a}\:\mathrm{second}\:\mathrm{hand}\:\mathrm{tshirt}\:\mathrm{and}\:\mathrm{earn} \\ $$$$\mathrm{c}{o}\mathrm{mmission}\:\mathrm{of}\:\mathrm{9\%}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{total}\:\mathrm{sales}.\mathrm{Each}\:\mathrm{t}−\mathrm{shirt} \\ $$$$\mathrm{cost}\:\mathrm{sh}\:\mathrm{7500}.\mathrm{Find}\:\mathrm{number}\:\mathrm{of}\:\mathrm{tshirt}\:\mathrm{sold}\:\mathrm{if} \\ $$$$\:\mathrm{mr}\:\mathrm{to}{n}\mathrm{y}\:\mathrm{earned}\:\mathrm{a}\:\mathrm{commission}\:\mathrm{of}\:\mathrm{297000} \\ $$ Terms of Service Privacy Policy…
Question Number 202522 by mou0113 last updated on 28/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202500 by Calculusboy last updated on 28/Dec/23 Answered by Frix last updated on 28/Dec/23 $$\mathrm{Obviously}\:{x}=\mathrm{1} \\ $$$$\sqrt[{\mathrm{2015}}]{\mathrm{1}+\mathrm{3}−\mathrm{3}}+\sqrt[{\mathrm{2015}}]{−\mathrm{1}−\mathrm{3}+\mathrm{5}}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$ Answered by Rasheed.Sindhi last…
Question Number 202497 by MATHEMATICSAM last updated on 28/Dec/23 $$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{qx}\:+\:{p}\:=\:\mathrm{0}\:\left[{p}\:\neq\:{q}\right]\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{p}\:+\:{q}\:+\:\mathrm{1}\:=\:\mathrm{0}. \\ $$ Answered by BaliramKumar…
Question Number 202530 by Calculusboy last updated on 28/Dec/23 Answered by MathematicalUser2357 last updated on 29/Dec/23 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{log}\left(\mathrm{1}+{x}\right)^{\mathrm{1}+{x}} −{x}}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\partial^{\mathrm{2}} }{\partial^{\mathrm{2}} {x}}\left(\mathrm{log}\left(\mathrm{1}+{x}\right)^{\mathrm{1}+{x}}…
Question Number 202514 by MrGHK last updated on 28/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202462 by pticantor last updated on 27/Dec/23 Answered by witcher3 last updated on 27/Dec/23 $$\mathrm{p}^{\mathrm{n}} −\mathrm{1}=\left(\mathrm{p}−\mathrm{1}\right)\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{k}} \right) \\ $$$$\Rightarrow\mathrm{p}.\left(\mathrm{p}^{\mathrm{n}} \right)+\left(\mathrm{p}−\mathrm{1}\right).\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}}…