Question Number 148483 by puissant last updated on 28/Jul/21 $$\mathrm{Soit}\:\mathrm{f}\:\mathrm{une}\:\mathrm{fonction}\:\mathrm{continu}\:\mathrm{sur}\:\mathbb{R} \\ $$$$\mathrm{et}\:\mathrm{non}\:\mathrm{identiquement}\:\mathrm{nulle}, \\ $$$$\forall\:\mathrm{x},\mathrm{x}'\in\mathbb{R},\:\mathrm{f}\left(\mathrm{x}−\mathrm{x}'\right)+\mathrm{f}\left(\mathrm{x}+\mathrm{x}'\right)=\mathrm{2f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{x}'\right) \\ $$$$\mathrm{montrer}\:\mathrm{que}: \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{1}\:\mathrm{et}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(−\mathrm{x}\right).. \\ $$ Answered by Olaf_Thorendsen last updated…
Question Number 17401 by ajfour last updated on 05/Jul/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{4}-\mathrm{digit}\:\mathrm{greatest} \\ $$$$\mathrm{number}\:\mathrm{and}\:\mathrm{the}\:\mathrm{5}-\mathrm{digit}\:\mathrm{smallest} \\ $$$$\mathrm{number},\:\mathrm{each}\:\mathrm{number}\:\mathrm{having}\:\mathrm{three} \\ $$$$\mathrm{different}\:\mathrm{digits}. \\ $$ Commented by RasheedSoomro last updated on 05/Jul/17…
Question Number 148398 by puissant last updated on 27/Jul/21 Answered by Olaf_Thorendsen last updated on 27/Jul/21 $$\mathrm{1}.\:\mathrm{C}_{\mathrm{1}} \:=\:\mathrm{7}{l}\:\mathrm{et}\:\mathrm{C}_{\mathrm{2}} \:=\:\mathrm{4}{l} \\ $$$$ \\ $$$$\bullet\:\mathrm{Etape}\:\mathrm{1}. \\ $$$$\mathrm{Je}\:\mathrm{remplis}\:\mathrm{C}_{\mathrm{1}}…
Question Number 82839 by M±th+et£s last updated on 24/Feb/20 $${if}\:{the}\:{first}\:{and}\:{fifth}\:{terms}\:{of}\:{arithmetic} \\ $$$${peogression}\:{are}\:{equal}\:{and}\:{the}\:{seventh} \\ $$$${and}\:{fourtenth}\:{terms}\:{of}\:{another}\:{arithmetic}\:{are} \\ $$$${equal}\:{then}\:{show}\:{that}\:{the}\:{first}\:{term}\:{from} \\ $$$${the}\:{first}\:{arithmetic}\:{is}\:{equal}\:{the}\:{tenth} \\ $$$${from}\:{the}\:{second}\:{one} \\ $$$${and}\:{so}\:{sorry}\:{because}\:{my}\:{english}\:{is} \\ $$$${not}\:{so}\:{good} \\…
Question Number 148368 by bouchrag last updated on 27/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 17280 by tawa tawa last updated on 03/Jul/17 $$\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{cosh}\left(\mathrm{2x}\right)\:=\:\mathrm{2cosh}^{\mathrm{2}} \left(\mathrm{x}\right)\:−\:\mathrm{1} \\ $$ Commented by mrW1 last updated on 03/Jul/17 $$\mathrm{cosh}\:\left(\mathrm{2x}\right)=\frac{\mathrm{e}^{\mathrm{2x}} +\mathrm{e}^{−\mathrm{2x}} }{\mathrm{2}} \\…
Question Number 148334 by Sravanth last updated on 27/Jul/21 Answered by Rasheed.Sindhi last updated on 27/Jul/21 $$\mathrm{3}^{\mathrm{6}} ×\left(\mathrm{2}^{−\mathrm{2}} ×\mathrm{3}^{\mathrm{5}×−\mathrm{2}} \right)×\left(\mathrm{2}×\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{3}^{\mathrm{6}} ×\mathrm{2}^{−\mathrm{2}} ×\mathrm{3}^{−\mathrm{10}}…
Question Number 17187 by virus last updated on 02/Jul/17 Answered by Tinkutara last updated on 02/Jul/17 $$\frac{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{2}} \:+\:…\:+\:{a}_{{p}} }{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{2}} \:+\:…\:+\:{a}_{{q}} }\:=\:\frac{\frac{{p}}{\mathrm{2}}\left[\mathrm{2}{a}_{\mathrm{1}} \:+\:\left({p}\:−\:\mathrm{1}\right){d}\right]}{\frac{{q}}{\mathrm{2}}\left[\mathrm{2}{a}_{\mathrm{1}} \:+\:\left({q}\:−\:\mathrm{1}\right){d}\right]}…
Question Number 148219 by iloveisrael last updated on 26/Jul/21 $$\mathrm{The}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+\mathrm{px}+\mathrm{qx}^{\mathrm{2}} \right)^{\mathrm{8}} \: \\ $$$$=\:\mathrm{1}+\mathrm{8x}+\mathrm{52x}^{\mathrm{2}} +\mathrm{kx}^{\mathrm{3}} +… \\ $$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{p}\:,\mathrm{q}\:\mathrm{and}\:\mathrm{k} \\ $$ Answered by liberty last updated…
Question Number 17102 by tawa tawa last updated on 30/Jun/17 $$\mathrm{compute}:\:\:\:\underset{\mathrm{k}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2k}\:+\:\mathrm{1}}{\mathrm{2}^{\mathrm{2}\left(\mathrm{k}\:+\:\mathrm{1}\right)} } \\ $$ Commented by prakash jain last updated on 01/Jul/17 $$\mathrm{S}=\:\underset{\mathrm{k}\:=\:\mathrm{0}}…