Question Number 22264 by kasiulka202 last updated on 14/Oct/17 $${What}\:{is}\:\mathrm{1}+\mathrm{5}\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Answered by Joel577 last updated on 14/Oct/17 $$\mathrm{U}\:\mathrm{have}\:\mathrm{1}\:\mathrm{cake},\:\mathrm{and}\:\mathrm{then}\:\mathrm{u}\:\mathrm{buy}\:\mathrm{5}\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{cakes} \\ $$$$\mathrm{So},\:\mathrm{u}\:\mathrm{have}\:\mathrm{total}\:\mathrm{6}\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{cakes} \\ $$ Terms…
Question Number 87647 by john santu last updated on 05/Apr/20 $$\mathrm{If}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:,\:\mathrm{a}_{\mathrm{n}+\mathrm{1}\:} =\:\mathrm{2a}_{\mathrm{n}} \:+\:\mathrm{5}\:,\:\mathrm{n}\:=\: \\ $$$$\mathrm{1},\mathrm{2},\mathrm{3},….\:\mathrm{then}\:\mathrm{a}_{\mathrm{100}} \:=\:? \\ $$ Answered by mr W last updated…
Question Number 22090 by tawa tawa last updated on 10/Oct/17 Answered by Tikufly last updated on 11/Oct/17 $$\left(\mathrm{i}\right)\:\mathrm{3},\:\:\mathrm{15},\:\:\:\mathrm{75},\:\:\:\mathrm{375},……. \\ $$$$\left(\mathrm{ii}\right)\mathrm{3},\:\:\mathrm{15},\:\:\:\mathrm{27},\:\:\:\mathrm{39},……… \\ $$ Commented by tawa…
Question Number 22076 by ajfour last updated on 10/Oct/17 $${find}\:{approximately}\:{and}\:{quickly} \\ $$$${without}\:{calculator}\:\frac{\mathrm{54329}}{\mathrm{2467}}\:. \\ $$ Commented by ajfour last updated on 10/Oct/17 $${my}\:{way}: \\ $$$$\:\frac{\mathrm{54329}}{\mathrm{2467}}\approx\frac{\mathrm{543}.\mathrm{3}}{\mathrm{24}.\mathrm{67}}\:\approx\:\frac{\mathrm{543}.\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{22}}{\mathrm{24}.\mathrm{67}+\frac{\mathrm{1}}{\mathrm{3}}} \\…
Question Number 87563 by Serlea last updated on 05/Apr/20 $$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{numbers} \\ $$$$\frac{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +…+\mathrm{2}^{\mathrm{n}−\mathrm{1}} }{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +…+\mathrm{2}^{\mathrm{n}} }\:\mathrm{and}\: \\ $$$$\frac{\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +…+\mathrm{3}^{\mathrm{n}−\mathrm{1}} }{\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +…+\mathrm{3}^{\mathrm{n}}…
Question Number 22023 by $@ty@m last updated on 09/Oct/17 Commented by $@ty@m last updated on 09/Oct/17 $${Solve}\:{this}\:{question}\:{in}\:{least}\:{steps}. \\ $$ Commented by Tinkutara last updated on…
Question Number 153070 by bobhans last updated on 04/Sep/21 $$\:\:\:{Find}\:{the}\:{solution}\:{set}\:{of}\:{inequality} \\ $$$$\:\:\:\:\:\:\:\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\geqslant\:{x}−\mathrm{2}\: \\ $$ Answered by Rasheed.Sindhi last updated on 04/Sep/21 $$\:\:\:\:\:\:\:\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\geqslant\:{x}−\mathrm{2}\: \\ $$$$\:\:\:\:\:\:\:\left(\:\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\geqslant\:{x}−\mathrm{2}\:\right)^{\mathrm{2}} \\…
Question Number 87322 by peter frank last updated on 04/Apr/20 $$\mathrm{I}{nvestigate}\:{the}\:{stationary} \\ $$$${value}\:{of} \\ $$$$\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{and}\:{sketch}\:{the}\:{graph} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 21738 by Joel577 last updated on 02/Oct/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{from}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{1}\:.\:\mathrm{2}^{\mathrm{1}} \:+\:\mathrm{2}\:.\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{3}\:.\:\mathrm{2}^{\mathrm{3}} \:+\:…\:+\:\mathrm{50}\:.\:\mathrm{2}^{\mathrm{50}} \:? \\ $$ Answered by $@ty@m last updated on 02/Oct/17…
Question Number 21732 by tawa tawa last updated on 02/Oct/17 $$\left(\mathrm{i}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{three}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{2}\:−\:\mathrm{x}\right)^{\mathrm{6}} \:\mathrm{in}\:\mathrm{ascending}\:\mathrm{power} \\ $$$$\mathrm{of}\:\mathrm{x}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{for}\:\mathrm{which}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{term}\:\mathrm{in}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\left(\mathrm{1}\:+\:\mathrm{kx}\right)\left(\mathrm{2}\:−\:\mathrm{x}\right)^{\mathrm{6}} \\ $$ Commented by Tikufly last…