Question Number 178810 by Ib last updated on 21/Oct/22 Commented by ARUNG_Brandon_MBU last updated on 21/Oct/22 Tu peux essayer de rogner l'image la prochaine fois pour améliorer la qualité car là c'est pas si agréable pour la vue. Answered by ARUNG_Brandon_MBU last updated on 21/Oct/22 $${z}=\frac{\mathrm{1}−{e}^{{i}\frac{\pi}{\mathrm{3}}}…
Question Number 178793 by infinityaction last updated on 21/Oct/22 $$\mathrm{find}\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{3}{n}\right)!}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{6}!}+\frac{\mathrm{1}}{\mathrm{9}!}+… \\ $$ Answered by aleks041103 last updated on 21/Oct/22 $${w}={e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} \Rightarrow{w}^{\mathrm{2}} ={w}^{\ast} ,{w}^{\mathrm{3}}…
Question Number 178794 by kouameMarcellin last updated on 21/Oct/22 $${i} \\ $$$${i} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 47712 by Rio Michael last updated on 13/Nov/18 $${show}\:{that}\: \\ $$$$\:\:^{\:^{\mathrm{2}} \:{C}_{\mathrm{2}} \:} {C}_{{n}} =\:\frac{\mathrm{1}}{\left(\mathrm{1}−{n}\right)!\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)…\mathrm{3}\left(\mathrm{2}\right)\left(\mathrm{1}\right)} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 178777 by mr W last updated on 21/Oct/22 $${A}=\frac{\mathrm{1}}{\mathrm{2018}}+\frac{\mathrm{1}}{\mathrm{2019}}+…+\frac{\mathrm{1}}{\mathrm{2050}} \\ $$$${find}\:{the}\:{integer}\:{part}\:{of}\:\frac{\mathrm{1}}{{A}}. \\ $$ Commented by Frix last updated on 21/Oct/22 $$\frac{\mathrm{1}}{{A}}\:\mathrm{should}\:\mathrm{be}\:\mathrm{close}\:\mathrm{to}\:\frac{\mathrm{2018}+\mathrm{2050}}{\mathrm{2}×\mathrm{33}}=\mathrm{61}.\mathrm{6363}… \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{61}…
Question Number 113241 by pallob last updated on 11/Sep/20 Answered by Aina Samuel Temidayo last updated on 11/Sep/20 $$\sqrt[{\mathrm{5}}]{\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{30}} }\:}}=\sqrt[{\left(\mathrm{5}×\mathrm{2}×\mathrm{3}\right)}]{\mathrm{x}^{\mathrm{30}} }=\sqrt[{\mathrm{30}}]{\mathrm{x}^{\mathrm{30}} }=\mathrm{x}^{\frac{\mathrm{30}}{\mathrm{30}}} =\mathrm{x}^{\mathrm{1}} =\mathrm{x} \\…
Question Number 178747 by cortano1 last updated on 21/Oct/22 $$\:\:\begin{cases}{\mathrm{3x}=\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\mathrm{3x}=\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:}\\{\mathrm{3x}=\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\end{cases} \\ $$$$\:\mathrm{x}=? \\ $$ Answered by Rasheed.Sindhi last updated on 21/Oct/22 $$\:\:\begin{cases}{\mathrm{3x}=\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)….\left(\mathrm{i}\right)}\\{\mathrm{3x}=\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right)….\left(\mathrm{ii}\right)\:}\\{\mathrm{3x}=\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{11}\right)….\left(\mathrm{iii}\right)}\end{cases}\:;\:\mathrm{x}=? \\ $$$$\left({i}\right)\Rightarrow\mathrm{3}{x}\equiv\mathrm{2}+\mathrm{2}×\mathrm{5}\left({mod}\:\mathrm{5}\right) \\…
Question Number 113188 by mathdave last updated on 11/Sep/20 $${proporsed}\:{by}\:{m}.{n}\:{july}\:\mathrm{1790} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx} \\ $$ Answered by mathdave last updated on 11/Sep/20 $${my}\:{solution}\:{to} \\…
Question Number 178715 by CrispyXYZ last updated on 20/Oct/22 $$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{2}\leqslant{ax}^{\mathrm{2}} +{bx}+{c}\leqslant\mathrm{3}\:\mathrm{is}\:\left[\mathrm{2},\:\mathrm{3}\right] \\ $$$$\left.\mathrm{1}\right)\:\mathrm{if}\:{a}>\mathrm{0},\:{ax}^{\mathrm{2}} +\left({b}−\mathrm{3}\right){x}−{c}\leqslant\mathrm{0}\:\mathrm{has}\:\mathrm{and}\:\mathrm{only}\:\mathrm{has}\:\mathrm{10}\:\mathrm{integer}\:\mathrm{solutions}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:{x}:\:{ax}^{\mathrm{2}} +\left({b}−\mathrm{1}\right){x}+\mathrm{5}<\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 178697 by infinityaction last updated on 20/Oct/22 Commented by mr W last updated on 21/Oct/22 $${for}\:{a},{b},{c}\:\in\mathbb{R},\:{i}\:{think} \\ $$$${a}+{b}+{c}\leqslant\mathrm{2}\sqrt{\mathrm{3}} \\ $$ Commented by mr…