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Author: Tinku Tara

U-n-2-3-5-6-7-8-10-11-12-13-14-find-U-n-

Question Number 7562 by Master Moon last updated on 03/Sep/16 $$\left({U}_{{n}} \right)=\:\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{10},\mathrm{11},\mathrm{12},\mathrm{13},\mathrm{14},…. \\ $$$${find}\:{U}_{{n}} \\ $$ Commented by Rasheed Soomro last updated on 03/Sep/16 $${Perfect}\:{square}\:{numbers}\:{have}\:\:{been}\:{excluded}\:{from}…

Question-138631

Question Number 138631 by physicstutes last updated on 15/Apr/21 Commented by physicstutes last updated on 15/Apr/21 $$\mathrm{From}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{above}\:\mathrm{show}\:\mathrm{completely}\:\mathrm{how}\:\mathrm{the}\:\mathrm{force}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{pulley}\:\mathrm{can}\:\mathrm{be}\:\mathrm{obtained}. \\ $$ Commented by Dwaipayan Shikari…

x-2-16-x-find-x-

Question Number 138627 by KwesiDerek last updated on 15/Apr/21 $$\boldsymbol{\mathrm{x}}^{\mathrm{2}} =\mathrm{16}^{\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$ Commented by soudo last updated on 15/Apr/21 $$\mathrm{nice} \\ $$…

4-5-

Question Number 7552 by Master Moon last updated on 03/Sep/16 $$\sqrt{\mathrm{4}}+\mathrm{5}\:= \\ $$ Answered by FilupSmith last updated on 03/Sep/16 $$\sqrt{\mathrm{4}}=\pm\mathrm{2} \\ $$$$\therefore\sqrt{\mathrm{4}}+\mathrm{5}=\mathrm{5}\pm\mathrm{2}=\mathrm{3},\:\mathrm{7} \\ $$…

Question-138621

Question Number 138621 by mnjuly1970 last updated on 15/Apr/21 Answered by Dwaipayan Shikari last updated on 15/Apr/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma^{\mathrm{2}} \left({n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{2}{n}+\mathrm{2}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \Sigma{x}^{{n}} \left(\mathrm{1}−{x}\right)^{{n}} {dx}…

0-e-x-3-1-1-x-3-3-dx-x-1-3-1-3-

Question Number 138619 by Dwaipayan Shikari last updated on 15/Apr/21 $$\int_{\mathrm{0}} ^{\infty} \left({e}^{−{x}^{\sqrt{\mathrm{3}}} } −\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\sqrt{\mathrm{3}}} \right)^{\sqrt{\mathrm{3}}} }\right)\frac{{dx}}{{x}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\psi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$ Commented by mnjuly1970 last updated on…