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 $$\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}=\sqrt{{x}}+\sqrt{{y}} \\ $$$$\mathrm{3}+\sqrt{\mathrm{2}}={x}+{y}+\mathrm{2}\sqrt{{xy}} \\…
Question Number 70768 by abdusalamyussif@gmail.com last updated on 07/Oct/19 Commented by abdusalamyussif@gmail.com last updated on 07/Oct/19 $$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$ Answered by MJS last updated on…
Question Number 136283 by JulioCesar last updated on 20/Mar/21 Answered by Ar Brandon last updated on 20/Mar/21 $$\int\mathrm{e}^{\mathrm{g}\left(\mathrm{x}\right)} \left[\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}'\left(\mathrm{x}\right)+\mathrm{f}\:'\left(\mathrm{x}\right)\right]\mathrm{dx}=\mathrm{e}^{\mathrm{g}\left(\mathrm{x}\right)} \mathrm{f}\left(\mathrm{x}\right) \\ $$ Terms of Service…
Question Number 5203 by FilupSmith last updated on 30/Apr/16 $$\mathrm{Silly}\:\mathrm{question} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\frac{{a}}{{b}}}}×\frac{\mathrm{1}}{{b}} \\ $$$$ \\ $$$$\mathrm{Does}\:\mathrm{this}\:\mathrm{equal}: \\ $$$$\frac{\mathrm{1}}{{b}\sqrt{\frac{{a}}{{b}}}}=\frac{\mathrm{1}}{\:\sqrt{{b}^{\mathrm{2}} \frac{{a}}{{b}}}}=\frac{\mathrm{1}}{\:\sqrt{{ab}}}\:???? \\ $$$$ \\ $$$$\mathrm{Ive}\:\mathrm{never}\:\mathrm{done}\:\mathrm{this}\:\mathrm{so}\:\mathrm{it}\:\mathrm{has}\:\mathrm{oddly}\:\mathrm{surprised}\:\mathrm{me}…
Question Number 70679 by cesar.marval.larez@gmail.com last updated on 06/Oct/19 $${Hi},\:{i}\:{wanna}\:{learn}\:{more}\:{english}. \\ $$$${can}\:{someone}\:{help}\:{me}? \\ $$$${write}\:{please}\:+\mathrm{584249229498} \\ $$ Commented by mr W last updated on 07/Oct/19 $${here}\:{is}\:{definitively}\:{not}\:{the}\:{best}\:{place}\:{to}…
Question Number 70659 by aliesam last updated on 06/Oct/19 $${find}\:{the}\:{range}\:{algrbraically} \\ $$$$ \\ $$$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$ Answered by MJS last updated on 06/Oct/19 $$\sqrt{{x}^{\mathrm{2}}…