Question Number 142035 by Dwaipayan Shikari last updated on 25/May/21 $$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}/\mathrm{4}} }.\frac{\mathrm{3}^{\mathrm{1}/\mathrm{9}} }{\mathrm{4}^{\mathrm{1}/\mathrm{16}} }.\frac{\mathrm{5}^{\mathrm{1}/\mathrm{25}} }{\mathrm{6}^{\mathrm{1}/\mathrm{36}} }.\frac{\mathrm{7}^{\mathrm{1}/\mathrm{49}} }{\mathrm{8}^{\mathrm{1}/\mathrm{64}} }…={exp}\left(−\frac{\zeta'\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{log}\:\left(\mathrm{2}\right)\right) \\ $$ Answered by mindispower last…
Question Number 10914 by Saham last updated on 02/Mar/17 $$\mathrm{A}\:\mathrm{200}\:\mathrm{N}\:\mathrm{force}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{40}°\:\mathrm{above}\:\mathrm{the}\:\mathrm{horizontal}\:,\:\mathrm{drag}\:\mathrm{load}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{floor}.\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{kinetic}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{load}\:\mathrm{is}\:\mathrm{0}.\mathrm{30}\: \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{load}\:\mathrm{experiences}\:\mathrm{an}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{1}.\mathrm{2}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} , \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{load}. \\ $$ Answered by sandy_suhendra last updated on…
Question Number 10908 by ketto last updated on 01/Mar/17 Answered by geovane10math last updated on 01/Mar/17 $${DE}\:\mathrm{must}\:\mathrm{be}\://\:\mathrm{with}\:{BC} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 10900 by Saham last updated on 01/Mar/17 Commented by sandy_suhendra last updated on 02/Mar/17 $$\mathrm{the}\:\mathrm{picture}\:\mathrm{isn}'\mathrm{t}\:\mathrm{clear}\:\mathrm{and}\:\mathrm{too}\:\mathrm{small} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 10898 by Saham last updated on 01/Mar/17 Answered by sandy_suhendra last updated on 02/Mar/17 $$\mathrm{E}^{\mathrm{2}} =\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} +\left(\mathrm{pc}\right)^{\mathrm{n}} \\ $$$$\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} =\mathrm{E}^{\mathrm{2}} −\left(\mathrm{pc}\right)^{\mathrm{n}}…
Question Number 10899 by Saham last updated on 01/Mar/17 Answered by sandy_suhendra last updated on 02/Mar/17 Commented by sandy_suhendra last updated on 02/Mar/17 $$\mathrm{q}_{\mathrm{1}} =\mathrm{q}_{\mathrm{2}}…
Question Number 76404 by mind is power last updated on 27/Dec/19 $$\mathrm{Hello}\:\mathrm{have}\:\mathrm{nice}\:\mathrm{end}\:\mathrm{of}\:\mathrm{year}\:\mathrm{good}\:\mathrm{bless}\:\mathrm{you} \\ $$$$\mathrm{all}\:\mathrm{i}\:\mathrm{respond}\:\mathrm{note}\:\mathrm{in}\:\mathrm{y}\:\mathrm{re}\:\mathrm{message}\:\mathrm{becsuse}\:\mathrm{i}\:\mathrm{have}\:\mathrm{so}\:\mathrm{many}\:\mathrm{problemes} \\ $$$$\mathrm{that}\:\mathrm{mack}\:\mathrm{me}\:\mathrm{feel}\:\mathrm{no}\:\mathrm{pleasur}\:\mathrm{any}\:\mathrm{more}\:\mathrm{to}\:\mathrm{do}\:\mathrm{somthing} \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{its}\:\mathrm{importante}\:\mathrm{to}\:\mathrm{say}\:\mathrm{it}\:\mathrm{i}\:\mathrm{will}\:\mathrm{back}\:\mathrm{Soon}\:\mathrm{i}\:\mathrm{hop}\:\mathrm{so}\:\:\mathrm{Sorry}\:\mathrm{for} \\ $$$$\mathrm{my}\:\mathrm{English} \\ $$ Commented by mr…
Question Number 76384 by Dofynia last updated on 26/Dec/19 $$ \\ $$ Commented by john santu last updated on 27/Dec/19 $${what}\:{is}\:{mean}? \\ $$ Terms of…
Question Number 76368 by Rio Michael last updated on 26/Dec/19 $${prove}\:{that} \\ $$$$\mathrm{1}.\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{r}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{2}.\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{r}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{6}}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}\:+\:\mathrm{1}\right) \\ $$$$\mathrm{3}.\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{4}}{n}^{\mathrm{2}}…
Question Number 76367 by Rio Michael last updated on 26/Dec/19 $${prove}\:{that}\:\:\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$ Commented by mathmax by abdo last updated on…