Question Number 46139 by pieroo last updated on 21/Oct/18 $$\mathrm{A}\:\mathrm{park}\:\mathrm{has}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{hexagon} \\ $$$$\mathrm{of}\:\mathrm{sides}\:\mathrm{2km}\:\mathrm{each}.\:\mathrm{A}\:\mathrm{boy}\:\mathrm{walks}\:\mathrm{a}\:\mathrm{distance} \\ $$$$\mathrm{of}\:\mathrm{5km}\:\mathrm{along}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{park}.\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{direct}\:\mathrm{distance}\:\mathrm{between}\: \\ $$$$\mathrm{the}\:\mathrm{start}\:\mathrm{point}\:\mathrm{and}\:\mathrm{the}\:\mathrm{end}\:\mathrm{point}? \\ $$ Commented by malwaan last updated…
Question Number 177193 by peter frank last updated on 02/Oct/22 $$\int\:\:\:\frac{\mathrm{x}^{\mathrm{2021}} }{\mathrm{x}^{\mathrm{2022}} +\mathrm{x}^{\mathrm{4043}} }\mathrm{dx} \\ $$ Answered by cortano1 last updated on 02/Oct/22 $$\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2021}} +\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{x}}\:−\frac{\mathrm{x}^{\mathrm{2020}}…
Question Number 46110 by Tawa1 last updated on 21/Oct/18 Commented by math khazana by abdo last updated on 22/Oct/18 $$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow\alpha=\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{and}\:\beta=\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\alpha={e}^{{i}\frac{\pi}{\mathrm{3}}} \:{and}\:\beta\:={e}^{−{i}\frac{\pi}{\mathrm{3}}} \:\Rightarrow\alpha^{\mathrm{101}}…
Question Number 177176 by mr W last updated on 01/Oct/22 $${solve} \\ $$$$\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }+\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }=\mathrm{13} \\ $$ Answered by a.lgnaoui last updated on 02/Oct/22 $${posons}\:\:{x}^{\mathrm{3}}…
Question Number 177173 by Shrinava last updated on 01/Oct/22 Answered by Peace last updated on 02/Oct/22 $${let}\:{f}\left({x}\right)={e}^{−\frac{\mathrm{1}}{{x}+\mathrm{1}}} ,{applie}\:{Taylors}\:{lagrange}\:{formul}\:\Rightarrow\exists{c}\in\left[\mathrm{0},\mathrm{1}\right]\:{such}\:{that} \\ $$$${f}\left({x}\right)={f}\left(\mathrm{0}\right)+{xf}'\left({c}\right)={e}^{−\mathrm{1}} +{x}.\frac{{e}^{−\frac{\mathrm{1}}{\mathrm{1}+{c}}} }{\left(\mathrm{1}+{c}\right)^{\mathrm{2}} } \\ $$$${We}\:{have}\:\forall{c}\in\left[\mathrm{0},\mathrm{1}\right]\:\frac{{e}^{−\frac{\mathrm{1}}{\mathrm{1}+{c}}}…
Question Number 46088 by Cheyboy last updated on 21/Oct/18 Answered by $@ty@m last updated on 21/Oct/18 $${Let}\:{Solution}\:{set}\:=\begin{bmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{bmatrix} \\ $$$${a}+{b}=\mathrm{8}\:…\left(\mathrm{1}\right) \\ $$$${b}+{d}=\mathrm{8}\:…\left(\mathrm{2}\right) \\ $$$$\Rightarrow{a}={d}\:…\left(\mathrm{3}\right) \\ $$$${c}−{d}=\mathrm{6}\:…\left(\mathrm{4}\right)…
Question Number 111623 by weltr last updated on 04/Sep/20 $$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$ Commented by ZiYangLee last updated on 04/Sep/20 $${x}\leqslant−\mathrm{3}\:\cup\:{x}\geqslant\mathrm{2}? \\ $$ Commented by…
Question Number 177144 by Shrinava last updated on 01/Oct/22 $$\mathrm{Calculas}: \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:…\:+\:\mathrm{16}^{\mathrm{2}} \:−\:\mathrm{16}}{\mathrm{1}\centerdot\mathrm{3}\:+\:\mathrm{2}\centerdot\mathrm{4}\:+\:\mathrm{3}\centerdot\mathrm{5}\:+\:…\:+\:\mathrm{15}\centerdot\mathrm{17}} \\ $$ Answered by BaliramKumar last updated on 01/Oct/22…
Question Number 111601 by A8;15: last updated on 04/Sep/20 Answered by bemath last updated on 04/Sep/20 $$\frac{\mathrm{7}.\mathrm{7}^{\mathrm{2004}} +\mathrm{7}^{\mathrm{2004}} +\mathrm{1}}{\mathrm{7}^{\mathrm{2004}} +\mathrm{1}}\:=\:\frac{\mathrm{8}.\mathrm{7}^{\mathrm{2004}} +\mathrm{1}}{\mathrm{7}^{\mathrm{2004}} +\mathrm{1}} \\ $$ Terms…
Question Number 111586 by Khanacademy last updated on 04/Sep/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{DTM}}−\mathrm{2020}\:\:\boldsymbol{{savollaridan}}\:\:\boldsymbol{{biri}}. \\ $$$$\:\int\boldsymbol{{x}}^{\boldsymbol{{x}}} =\:? \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{t}}.\boldsymbol{{me}}/\boldsymbol{{matematik\_olimpiadachilar}} \\ $$ Commented by Her_Majesty last updated on 04/Sep/20 $${see}\:{comment}\:{to}\:{question}\:\mathrm{111558}…