Question Number 136494 by liberty last updated on 22/Mar/21 $$\mathrm{5}\left(\sqrt{\mathrm{1}−{x}}\:+\sqrt{\mathrm{1}+{x}}\:\right)=\:\mathrm{6}{x}\:+\:\mathrm{8}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\: \\ $$ Answered by MJS_new last updated on 22/Mar/21 $$\mathrm{I}\:\mathrm{get}\:\mathrm{2}\:\mathrm{solutions} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{24}}{\mathrm{25}} \\…
Question Number 5414 by FilupSmith last updated on 14/May/16 $$\mathrm{How}\:\mathrm{do}\:\mathrm{we}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}/\mathrm{rectangle} \\ $$$$\mathrm{is}\:\mathrm{at}\:\mathrm{its}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$ \\ $$$$\mathrm{So},\:\mathrm{if}\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{of}\:\mathrm{sides}\:{a}\:\mathrm{and}\:{b}, \\ $$$$\mathrm{how}\:\mathrm{do}\:\mathrm{we}\:\mathrm{show}\:\mathrm{its}\:\mathrm{maximized}\:\mathrm{when}\:{a}={b}? \\ $$ Answered by 123456 last updated…
Question Number 136475 by EDWIN88 last updated on 22/Mar/21 Commented by EDWIN88 last updated on 22/Mar/21 $$\mathrm{Use}\:\mathrm{the}\:\mathrm{fact}\:\mathrm{that}\:\mathrm{csc}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{x}}{\mathrm{6}}+\frac{\mathrm{7x}^{\mathrm{3}} }{\mathrm{360}}+\frac{\mathrm{31x}^{\mathrm{5}} }{\mathrm{15120}}+…\:\mathrm{for}\:\mid\mathrm{x}\mid\:<\:\pi\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}\:\mathrm{for} \\ $$$$\mathrm{2}\:\mathrm{csc}\:\left(\mathrm{2x}\right)\:\left[\:\mathrm{for}\:\mid\mathrm{x}\mid\:<\:\frac{\pi}{\mathrm{2}}\:\mathrm{and}\:\mathrm{thus}\:\mathrm{for}\:\mathrm{ln}\:\left(\mathrm{tan}\:\mathrm{x}\right).\right. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{ln}\:\mid\mathrm{x}\mid\:+\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 70920 by TawaTawa last updated on 09/Oct/19 $$\mathrm{i}\:!\:\:=\:\:? \\ $$ Commented by prakash jain last updated on 10/Oct/19 $${i}!=\Gamma\left(\mathrm{1}+{i}\right) \\ $$$$\mathrm{where}\:{i}=\sqrt{−\mathrm{1}} \\ $$…
Question Number 5386 by sanusihammed last updated on 12/May/16 $${Solve}\:{simultaneously} \\ $$$$ \\ $$$${x}^{{x}} \:+\:\:{y}^{{y}\:} \:=\:\:\mathrm{31} \\ $$$${x}\:\:+\:\:{y}\:\:=\:\:\mathrm{5} \\ $$$$ \\ $$$${Please}\:{i}\:{did}\:{workings}\:… \\ $$$$ \\…
Question Number 5385 by sanusihammed last updated on 12/May/16 $${Find}\:{the}\:{value}\:{of}\:{x}\:{if}\: \\ $$$$ \\ $$$$\sqrt{{x}^{{x}^{\mathrm{6}} } }\:=\:\:\mathrm{729} \\ $$ Answered by prakash jain last updated on…
Question Number 136433 by metamorfose last updated on 21/Mar/21 $$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{\mathrm{2}{n}+\mathrm{1}} ^{\mathrm{2}{k}+\mathrm{1}} \mathrm{8}^{{k}} =…??? \\ $$ Answered by Olaf last updated on 22/Mar/21 $$…
Question Number 136408 by nimnim last updated on 21/Mar/21 $$\:\:\:\:\:\:{S}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{3}{k}^{\mathrm{2}} }{\mathrm{2}{k}^{\mathrm{3}} +\mathrm{2}}\:\:=? \\ $$$$\:\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}{k}^{\mathrm{2}} }{{k}^{\mathrm{3}} +\mathrm{1}}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{k}+\mathrm{1}}+\frac{\mathrm{2}{k}−\mathrm{1}}{{k}^{\mathrm{2}} −{k}+\mathrm{1}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:{I}\:{dont}\:{know}\:{how}\:{to}\:{continue}…{Please}\:{Help}.…
Question Number 5315 by FilupSmith last updated on 07/May/16 $$\mathrm{My}\:\mathrm{mother}\:\mathrm{is}\:\mathrm{making}\:\mathrm{a}\:\mathrm{game}\:\mathrm{for}\:\mathrm{children} \\ $$$$\mathrm{that}\:\mathrm{involves}\:\mathrm{words}\:\mathrm{on}\:\mathrm{cards}. \\ $$$$\mathrm{e}.\mathrm{g}.\:\mathrm{card}\:\mathrm{one}\:\mathrm{hase}\:\mathrm{words}\:{a}\:\mathrm{and}\:{b}, \\ $$$$\mathrm{for}\:\mathrm{whatever}\:{a}\:\mathrm{and}\:{b}\:\mathrm{might}\:\mathrm{be}. \\ $$$$ \\ $$$$\mathrm{She}\:\mathrm{has}\:\mathrm{a}\:\mathrm{list}\:\mathrm{of}\:\mathrm{20}\:\mathrm{words}\:\mathrm{and}\:\mathrm{she}\:\mathrm{wants}\:\mathrm{cards} \\ $$$$\mathrm{that}\:\mathrm{are}'\mathrm{t}\:\mathrm{duplicate}.\:\mathrm{That}\:\mathrm{is},\:{a}\:{b}\:\mathrm{and}\:{b}\:{a} \\ $$$$\mathrm{are}\:\mathrm{duplicates}. \\…
Question Number 136340 by liberty last updated on 21/Mar/21 $${If}\:\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}\:=\:\sqrt{\frac{{a}+\sqrt{{b}}}{{c}}}\:+\:\sqrt{\frac{{a}−\sqrt{{b}}}{{c}}} \\ $$$${then}\:{a}+{bc}\:=\:? \\ $$ Answered by Ñï= last updated on 21/Mar/21 $$\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}=\sqrt{{x}}+\sqrt{{y}} \\ $$$$\mathrm{3}+\sqrt{\mathrm{2}}={x}+{y}+\mathrm{2}\sqrt{{xy}} \\…