Question Number 166641 by pete last updated on 23/Feb/22 $$\mathrm{From}\:\mathrm{the}\:\mathrm{standard}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}, \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{origin}\:\left(\mathrm{0},\mathrm{0}\right),\:\mathrm{we}\:\mathrm{deduced}\:\mathrm{the}\:\mathrm{eqution} \\ $$$$\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} . \\ $$$$\mathrm{In}\:\mathrm{what}\:\mathrm{terms}\:\mathrm{do}\:\mathrm{we}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formular}? \\ $$ Commented…
Question Number 101104 by student work last updated on 30/Jun/20 $$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}!\right)=? \\ $$ Answered by smridha last updated on 30/Jun/20 $$\boldsymbol{{x}}!=\boldsymbol{\Gamma}\left(\boldsymbol{{x}}+\mathrm{1}\right)=\boldsymbol{{x}}\Gamma\boldsymbol{{x}} \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\boldsymbol{{n}}!}{\left(\boldsymbol{{x}}+\mathrm{1}\right)\left(\boldsymbol{{x}}+\mathrm{2}\right)……\left(\boldsymbol{{x}}+\boldsymbol{{n}}\right)}\boldsymbol{{n}}^{\boldsymbol{{x}}} \left[\boldsymbol{{E}}{u}\boldsymbol{{ler}}\:\boldsymbol{{definition}}\right]…
Question Number 166633 by mathlove last updated on 23/Feb/22 Answered by mahdipoor last updated on 23/Feb/22 $${xyz}={log}_{\mathrm{2}{a}} {a}×{log}_{\mathrm{3}{a}} \mathrm{2}{a}×{log}_{\mathrm{4}{a}} \mathrm{3}{a}= \\ $$$$\frac{{lna}}{{ln}\mathrm{2}{a}}×\frac{{ln}\mathrm{2}{a}}{{ln}\mathrm{3}{a}}×\frac{{ln}\mathrm{3}{a}}{{ln}\mathrm{4}{a}}=\frac{{lna}}{{ln}\mathrm{4}{a}} \\ $$$${yz}={log}_{\mathrm{3}{a}} \mathrm{2}{a}×{log}_{\mathrm{4}{a}}…
Question Number 101096 by student work last updated on 30/Jun/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{oblique}\:\mathrm{asymptote}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\centerdot\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \:\:\:\:\: \\ $$$$\:\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help} \\ $$ Commented by student work last updated on 30/Jun/20 $$\mathrm{i}\:\mathrm{need}\:\mathrm{soon}…
Question Number 166620 by mathlove last updated on 23/Feb/22 Answered by benhamimed last updated on 23/Feb/22 $${x}+\frac{\mathrm{1}}{{x}}\neq\sqrt{\mathrm{3}}\:\:\:\:\:\forall\:{x}\in\mathbb{R}^{\ast} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)'=\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\left.\left(\left.{x}+\frac{\mathrm{1}}{{x}}\right)\in\right]−\infty;−\mathrm{2}\right]\cup\left[\mathrm{2};+\infty\left[\right.\right. \\…
Question Number 101085 by ajfour last updated on 30/Jun/20 Commented by ajfour last updated on 30/Jun/20 $${x}\:{in}\:{terms}\:{of}\:{b}\:{and}\:{c},\:{given}\:{any} \\ $$$${real}\:{b}\:{and}\:{c}. \\ $$$$\left({what}\:{seems}\:{a}\:{right}\:{angle}\:{is}\:\right. \\ $$$$\left.{indeed}\:{a}\:{right}\:{angle}\right) \\ $$…
Question Number 166611 by cortano1 last updated on 23/Feb/22 $$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynom}\:\mathrm{give}\:\mathrm{the}\: \\ $$$$\:\mathrm{remainder}\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\: \\ $$$$\:\mathrm{x}^{\mathrm{4}} +\mathrm{4}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{divided}\:\mathrm{by}\:\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{2}. \\ $$ Terms of Service Privacy…
Question Number 101062 by bemath last updated on 30/Jun/20 Answered by MJS last updated on 30/Jun/20 $$\mathrm{after}\:“\mathrm{many}\:\mathrm{times}''\:\mathrm{implies}\:\mathrm{that} \\ $$$${p}=\frac{\mathrm{10}}{\mathrm{25}}=\frac{\mathrm{2}}{\mathrm{5}}\pm\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{there}\:\mathrm{are}\:\frac{\mathrm{25}}{{p}}=\mathrm{62}.\mathrm{5}\approx\mathrm{60}\:\mathrm{candies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{jar} \\ $$ Commented by…
Question Number 101056 by bemath last updated on 30/Jun/20 $$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{5}{x}+\mathrm{1}\right)+{p}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{one}\:\mathrm{root}\:\mathrm{equals}\:\mathrm{to}\:\mathrm{two}\:\mathrm{more} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{other},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${p}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\_\_\_ \\ $$ Commented by bemath last updated…
Question Number 166592 by Ari last updated on 22/Feb/22 Answered by nurtani last updated on 23/Feb/22 $${x}^{\mathrm{2}} =\mathrm{17}{x}+{y}…\left({i}\right) \\ $$$${y}^{\mathrm{2}} ={x}+\mathrm{17}{y}…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\blacksquare\:\:{x}^{\mathrm{2}}…