Question Number 143462 by Willson last updated on 14/Jun/21 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{{x}−\mathrm{1}}{{ln}\left(\frac{{x}}{\mathrm{2}−{x}}\right)}\:=\:??? \\ $$ Answered by Dwaipayan Shikari last updated on 14/Jun/21 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}−\mathrm{1}}{{log}\left(\frac{{x}}{\mathrm{2}−{x}}\right)}=\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{z}}{{log}\left(\frac{\mathrm{1}+{z}}{\mathrm{1}−{z}}\right)}=\frac{{z}}{\mathrm{2}\left({z}+\frac{{z}^{\mathrm{3}} }{\mathrm{3}}+..\right)}=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 12218 by tawa last updated on 16/Apr/17 $$\mathrm{5}!!\:=\:? \\ $$ Commented by tawa last updated on 16/Apr/17 $$\mathrm{How}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{15} \\ $$ Commented by prakash…
Question Number 12087 by tawa last updated on 12/Apr/17 $$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{10}\:\mathrm{objects}\:\mathrm{be}\:\mathrm{split}\:\mathrm{into}\:\mathrm{two}\:\:\mathrm{groups}\:\mathrm{containing}\: \\ $$$$\mathrm{4}\:\mathrm{and}\:\mathrm{6}\:\mathrm{objects}\:\mathrm{respectively}\:? \\ $$ Answered by sandy_suhendra last updated on 12/Apr/17 $$\mathrm{10C4}\:=\:\frac{\mathrm{10}!}{\mathrm{6}!\:\mathrm{4}!}\:=\mathrm{210} \\ $$$$\mathrm{or}\:\:\mathrm{10C6}\:=\:\mathrm{210} \\…
Question Number 77505 by jagoll last updated on 07/Jan/20 $${many}\:{three}\: \\ $$$${digit}\:{multiples}\:{of}\:{three}\:{that} \\ $$$${can}\:{be}\:{made}\:{from}\:{the}\:{number} \\ $$$$\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7}\:{without}\:{repetition}? \\ $$ Commented by jagoll last updated on 07/Jan/20…
Question Number 11855 by tawa last updated on 02/Apr/17 $$\mathrm{Ten}\:\mathrm{men}\:\mathrm{are}\:\mathrm{present}\:\mathrm{at}\:\mathrm{a}\:\mathrm{club}.\:\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{four}\:\mathrm{be}\:\mathrm{chosen}\:\mathrm{to} \\ $$$$\mathrm{play}\:\mathrm{bridge}\:\mathrm{if}\:\mathrm{two}\:\mathrm{men}\:\mathrm{refuse}\:\mathrm{to}\:\mathrm{sit}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{table}. \\ $$ Answered by sandy_suhendra last updated on 03/Apr/17 $$\mathrm{let}\:\mathrm{the}\:\mathrm{2}\:\mathrm{men}\:\mathrm{are}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B} \\ $$$$\mathrm{if}\:\mathrm{A}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{and}\:\mathrm{B}\:\mathrm{isn}'\mathrm{t}=\:\mathrm{8C3} \\…
Question Number 142904 by liberty last updated on 07/Jun/21 $${If}\:\mathrm{2}+\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\mathrm{3}+\mathrm{log}\:_{\mathrm{3}} \left({y}\right)=\mathrm{log}\:_{\mathrm{6}} \left({x}−\mathrm{4}{y}\right) \\ $$$${then}\:{the}\:{value}\:{of}\:\frac{\mathrm{1}}{\mathrm{2}{y}}−\frac{\mathrm{2}}{{x}}=? \\ $$ Answered by bramlexs22 last updated on 07/Jun/21 Terms…
Question Number 142765 by mr W last updated on 05/Jun/21 $${A}\:{class}\:{has}\:\mathrm{13}\:{children}.\:{To}\:{play}\:{a} \\ $$$${game}\:{one}\:{child}\:{is}\:{the}\:{referee}\:{and}\:{the} \\ $$$${other}\:{children}\:{are}\:{divided}\:{in}\:{three}\: \\ $$$${teams}\:{with}\:{four}\:{children}\:{in}\:{each}\:{team}. \\ $$$${In}\:{how}\:{many}\:{ways}\:{can}\:{the}\:{class}\:{play} \\ $$$${the}\:{game}? \\ $$ Commented by…
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Question Number 77199 by Boyka last updated on 04/Jan/20 Commented by Boyka last updated on 04/Jan/20 $$\left.\mathrm{A}\left.\right)\left.\mathrm{4}\left.\mathrm{05}\:\:\:\:\:\:\mathrm{B}\right)\mathrm{404}\:\:\:\:\:\:\:\mathrm{C}\right)\mathrm{403}\:\:\:\:\:\:\:\:\:\mathrm{D}\right)\mathrm{401} \\ $$ Commented by Boyka last updated on…
Question Number 142691 by SLVR last updated on 04/Jun/21 $${The}\:{number}\:{of}\:{distributions}\:{of}\:\mathrm{52} \\ $$$${cards}\:{divided}\:{equally}\:{to}\:\mathrm{4}\:{persons}\:{so} \\ $$$${as}\:{each}\:{gets}\:\mathrm{4}\:{cards}\:{of}\:{same}\:{suit} \\ $$$${taken}\:{away}\:{from}\:\mathrm{3}{suits}\left(\mathrm{4}×\mathrm{3}=\mathrm{12}\right)\wp \\ $$$${remaining}\:{card}\:{from}\:{remaining} \\ $$$$\mathrm{4}\:{th}\:{suit}\:{is} \\ $$ Commented by SLVR…