Question Number 129 by anoopsingh7374@gmail.com last updated on 25/Jan/15 $$\mathrm{3}{x}.\mathrm{6}{x}.\mathrm{7}{x}= \\ $$ Answered by rajabhay last updated on 07/Dec/14 $$\mathrm{3}{x}\centerdot\mathrm{6}{x}\centerdot\mathrm{7}{x}=\mathrm{126}{x}^{\mathrm{3}} \\ $$ Terms of Service…
Question Number 96 by sudhanshur last updated on 25/Jan/15 $${x}=\mathrm{17}^{\mathrm{4}} \:\mathrm{and}\:{y}=\mathrm{14}×\mathrm{16}×\mathrm{18}×\mathrm{20}\:\mathrm{then} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{greater}? \\ $$ Answered by anoopsingh7374@gmail.com last updated on 27/Nov/14 $$ \\ $$…
Question Number 93 by mreddy last updated on 25/Jan/15 $$\mathrm{Fnd}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\mathrm{n}+\mathrm{3}} −\mathrm{2}^{\mathrm{n}+\mathrm{1}} ×\mathrm{10}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} ×\mathrm{6}} \\ $$ Answered by sushmitak last updated on 28/Nov/14 $$\frac{\mathrm{2}^{{n}+\mathrm{1}}…
Question Number 131144 by SWPlaysMC last updated on 16/Dec/21 $${delete}\:{this}\:{question}\:{please}\:{because}\:{I}\:{can}'{t} \\ $$$${was}\:{just}\:{trying}\:{it}\:{out} \\ $$$${if}\:{x}=\mathrm{10}{y}^{\mathrm{2}} \\ $$$${then}\:{x}>{x}−\mathrm{14} \\ $$$${else}\:{if}\:{x}=\mathrm{11}{y}^{\mathrm{2}} \\ $$$${then}\:{x}<{x}+\mathrm{14} \\ $$$${end} \\ $$ Terms…
Question Number 131155 by EDWIN88 last updated on 02/Feb/21 $${Given}\:\begin{cases}{{a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}} }\\{{a}_{\mathrm{1}} =\mathrm{3}\:;\:{a}_{\mathrm{2}} =\mathrm{2}}\end{cases} \\ $$$$\:{find}\:{a}_{{n}} . \\ $$ Answered by john_santu last updated…
Question Number 131158 by EDWIN88 last updated on 02/Feb/21 $${If}\:\mathrm{4}{a}_{{n}} +\mathrm{2}{a}_{−{n}} =\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{3} \\ $$$${find}\:{a}_{{n}} \:=? \\ $$ Answered by mr W last updated on…
Question Number 71 by mike last updated on 25/Jan/15 $$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\ldots}}}=? \\ $$ Answered by jayant last updated on 17/Nov/14 $$\mathrm{1} \\ $$ Commented by 123456…
Question Number 65569 by MJS last updated on 31/Jul/19 $$\mathrm{reposting}\:\mathrm{this}: \\ $$$${x}^{\mathrm{4}} +\left(\mathrm{1}−\mathrm{2}{a}\right){x}^{\mathrm{2}} −\mathrm{2}{ax}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{a}>\mathrm{0}\wedge{x}>\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{5}} −\frac{\mathrm{23}}{\mathrm{8}}{x}^{\mathrm{4}} −\frac{\mathrm{49}}{\mathrm{4}}{x}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{8}}{x}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{8}}=\mathrm{0} \\…
Question Number 25 by user1 last updated on 25/Jan/15 $$\mathrm{Expand}\:\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} \\ $$ Answered by user1 last updated on 31/Oct/14 $$\:\left({a}+{b}+{c}\right)\left({a}+{b}+{c}\right) \\ $$$$={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca}…
Question Number 131084 by greg_ed last updated on 01/Feb/21 $$\mathrm{let}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{in}\:\mathbb{R}\::\: \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}. \\ $$ Answered by liberty last updated on 01/Feb/21 $$\mathrm{x}_{\mathrm{1}}…