Question Number 588 by shaleen last updated on 03/Feb/15 $$\left({a}+{b}=\mathrm{10}\right)\:\left({a}+{c}=\mathrm{24}\right)\:\left({b}+{c}=\mathrm{20}\right) \\ $$$${find}\:{a}\:{b}\:{c} \\ $$ Answered by prakash jain last updated on 03/Feb/15 $${a}+{b}=\mathrm{10}\Rightarrow{b}=\mathrm{10}−{a} \\ $$$${a}+{c}=\mathrm{24}\Rightarrow{c}=\mathrm{24}−{a}…
Question Number 577 by Bek last updated on 31/Jan/15 $${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$ Answered by Bek last updated on 31/Jan/15 $$ \\ $$…
Question Number 131640 by Salman_Abir last updated on 07/Feb/21 Answered by EDWIN88 last updated on 07/Feb/21 $$ \\ $$$$\mathrm{Du}\:\mathrm{bist}\:\mathrm{wundersch}\ddot {\mathrm{o}n} \\ $$ Commented by EDWIN88…
Question Number 131605 by physicstutes last updated on 06/Feb/21 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:{m}^{\mathrm{4}} −\mathrm{7}{m}^{\mathrm{3}} +\:\mathrm{14}{m}^{\mathrm{2}} −\mathrm{7}{m}\:+\:\mathrm{1}\:=\:\mathrm{0}\: \\ $$ Answered by malwan last updated on 06/Feb/21 $${m}^{\mathrm{4}}…
Question Number 482 by prakash jain last updated on 12/Jan/15 $$\mathrm{The}\:\mathrm{number}\:\mathrm{1000}!\:\mathrm{has}\:\mathrm{certain}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{0}{s}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end},\:\mathrm{what}\:\mathrm{the}\:\mathrm{the}\:\mathrm{first}\:\mathrm{non}−\mathrm{zero} \\ $$$$\mathrm{digit}. \\ $$$$\mathrm{1000}!=…\mathrm{d}_{\mathrm{1}} \mathrm{d}_{\mathrm{2}} \mathrm{d}_{\mathrm{3}} \mathrm{D00000}… \\ $$$$\mathrm{where}\:\mathrm{d}_{\mathrm{1}} ,\mathrm{d}_{\mathrm{2}} ,\mathrm{d}_{\mathrm{3}} ,\mathrm{D}\:\mathrm{are}\:\mathrm{digits}.…
Question Number 66004 by salahahmed last updated on 07/Aug/19 $${x}^{{x}^{{x}^{\iddots} } } =\mathrm{3} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x} \\ $$ Answered by mr W last updated on 07/Aug/19…
Question Number 65981 by Tanmay chaudhury last updated on 07/Aug/19 Answered by jimful last updated on 07/Aug/19 $${let}\:{s}_{{n}} =\Sigma\mathrm{1}/{n}. \\ $$$$\Sigma\left({k}+\mathrm{1}\right)/{k}\:\bullet\Sigma{k}/\left({k}+\mathrm{1}\right) \\ $$$$=\left({n}+{s}_{{n}} \right)\left({n}−{s}_{{n}+\mathrm{1}} +\mathrm{1}\right)…