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Question Number 194506 by Biteshwar last updated on 08/Jul/23 $$\:\:\:\:\:\:\:\:\:\:\mathrm{Gyanashram}\:\mathrm{classes} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{shivkund}\left(\mathrm{Munger}\right)\:\:\:\:\mathrm{by}−\mathrm{Bittu}\:\mathrm{sir} \\ $$$$\:\:\:\mathrm{12}\:\:\mathrm{th}\:\mathrm{physics}\:\mathrm{test} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{1}. \: \: \: \: \: \: \: \: \\…
Question Number 194500 by qaz last updated on 08/Jul/23 $$\underset{\mathrm{0}\leqslant{i}^{\mathrm{2}} +{j}^{\mathrm{2}} \leqslant\mathrm{16}} {\sum}\left({i}+{j}\right)=? \\ $$ Answered by MM42 last updated on 08/Jul/23 $$ \\ $$$${ans}=\mathrm{0}…
Question Number 194498 by SaRahAli last updated on 08/Jul/23 Commented by MM42 last updated on 08/Jul/23 $$? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 194445 by tri26112004 last updated on 06/Jul/23 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+{a}\right)}=¿ \\ $$$$\left({a}\neq\mathrm{0}\right) \\ $$ Answered by mr W last updated on 08/Jul/23…
Question Number 194436 by sonukgindia last updated on 06/Jul/23 Answered by Frix last updated on 06/Jul/23 $$\mathrm{3}\:?\:\sqrt{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$\mathrm{3}−\sqrt{\mathrm{3}}\:?\:\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$\mathrm{54}−\mathrm{30}\sqrt{\mathrm{3}}\:?\:\mathrm{2} \\ $$$$\mathrm{52}\:?\:\mathrm{30}\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{26}}{\mathrm{15}}\:?\:\sqrt{\mathrm{3}}…
Question Number 194434 by SANOGO last updated on 06/Jul/23 $${calcul} \\ $$$${e}^{\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)\:} −{e}^{−\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)} \:=? \\ $$ Answered by aba last updated on 06/Jul/23 $$\mathrm{e}^{\mathrm{2ln}\left(\mathrm{1}+\mathrm{u}\right)} −\mathrm{e}^{−\mathrm{2ln}\left(\mathrm{1}+\mathrm{u}\right)}…
Question Number 194363 by sonukgindia last updated on 05/Jul/23 Commented by Frix last updated on 05/Jul/23 $$\mathrm{There}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}. \\ $$$$\sqrt{{z}_{\mathrm{1}} +\sqrt{{z}_{\mathrm{2}} }}+\sqrt{{z}_{\mathrm{1}} −\sqrt{{z}_{\mathrm{2}} }}={r} \\ $$$$\mathrm{Solving}\:\mathrm{by}\:\mathrm{2}\:\mathrm{times}\:\left[\mathrm{squaring}\:\left(\mathrm{introduces}\right.\right.…
Question Number 194359 by Tinku Tara last updated on 05/Jul/23 $${v}\mathrm{2}.\mathrm{282}\:\mathrm{has}\:\mathrm{been}\:\mathrm{published}.\:\mathrm{This} \\ $$$$\mathrm{update}\:\mathrm{fixes}\:\mathrm{issues}\:\mathrm{with}\:\mathrm{missed} \\ $$$$\mathrm{notifications}\:\mathrm{and}\:\mathrm{adds}\:\mathrm{an} \\ $$$$\mathrm{arbitarary}\:\mathrm{precision}\: \\ $$$$\mathrm{scientific}\:\mathrm{calculator}. \\ $$ Commented by som(math1967) last…