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64-1-6-Is-there-any-short-cut-for-mcq-

Question Number 71326 by sadimuhmud 136 last updated on 13/Oct/19 $$\left(−\mathrm{64}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} =?\left(\boldsymbol{\mathrm{I}}\mathrm{s}\:\mathrm{there}\:\mathrm{any}\:\mathrm{short}\:\mathrm{cut}\:\mathrm{for}\:\mathrm{mcq}\right) \\ $$ Answered by MJS last updated on 13/Oct/19 $$\mathrm{2}^{\mathrm{6}} =\mathrm{64}\:\Rightarrow\:\sqrt[{\mathrm{6}}]{−\mathrm{64}}=\mathrm{2i} \\ $$…

n-a-n-n-1-2-b-n-n-1-2-1-2a-2-pi-0-e-t-2-2-1-a-ab-cosh-log-ab-t-a-3-b-2a-ab-cosh-log-ab-t-dt-

Question Number 136850 by Dwaipayan Shikari last updated on 26/Mar/21 $$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{a}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} {b}^{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} =\mathrm{1}+\sqrt{\frac{\mathrm{2}{a}^{\mathrm{2}} }{\pi}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} /\mathrm{2}} \left(\frac{\mathrm{1}−{a}\sqrt{{ab}}\:{cosh}\left(\sqrt{{log}\left({ab}\right)}\:{t}\right)}{{a}^{\mathrm{3}} {b}−\mathrm{2}{a}\sqrt{{ab}\:}\:{cosh}\left(\sqrt{{log}\left({ab}\right)}\:{t}\right)}\right){dt} \\ $$ Commented by…

1-1-2-2-1-2-1-1-3-2-2-2-1-2-2-2-1-3-5-2-3-1-2-3-3-1-4-2pi-3-1-4-2-

Question Number 136757 by Dwaipayan Shikari last updated on 25/Mar/21 $$\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}!}+\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }\right).\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}+….=\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{2}\pi^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{4}} }\right)^{\mathrm{2}} \\ $$ Answered by mindispower last…

Question-136760

Question Number 136760 by I want to learn more last updated on 25/Mar/21 Answered by mr W last updated on 25/Mar/21 $$\frac{{BC}}{\mathrm{sin}\:\angle{BDC}}=\frac{{CD}}{\mathrm{sin}\:\mathrm{30}°}\:\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\angle{BDA}}{{AB}}=\frac{\mathrm{sin}\:\mathrm{45}°}{{DA}}\:\:\:…\left({ii}\right) \\…