Category: Others

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 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 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 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 10807 by Saham last updated on 25/Feb/17 Commented by ridwan balatif last updated on 26/Feb/17 $$\mathrm{19}.\mathrm{13}\:\:\epsilon=\mathrm{1}\left(\mathrm{because}\:\mathrm{like}\:\mathrm{act}\:\mathrm{blackbody}\right),\:\mathrm{A}=\mathrm{1cm}^{\mathrm{2}} =\mathrm{10}^{−\mathrm{4}} \mathrm{m}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sigma\left(\mathrm{Boltzman}\:\mathrm{constant}\right)=\mathrm{5}.\mathrm{67}×\mathrm{10}^{−\mathrm{8}} \mathrm{Wm}^{−\mathrm{2}} \mathrm{K}^{−\mathrm{4}} ,\:\mathrm{T}=\mathrm{1727}^{\mathrm{o}}…