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 $$\mathrm{2}^{\mathrm{6}} =\mathrm{64}\:\Rightarrow\:\sqrt[{\mathrm{6}}]{−\mathrm{64}}=\mathrm{2i} \\ $$…
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…
Question Number 136826 by NancyJerotich last updated on 26/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 136799 by Dwaipayan Shikari last updated on 26/Mar/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{q}^{{n}} \right){dq} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 71235 by Rio Michael last updated on 13/Oct/19 $${sinh}\left[{ln}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\:\right]\:\equiv\: \\ $$$$ \\ $$$${A}.\:\:\mathrm{2}{x} \\ $$$${B}.\:\:\frac{\mathrm{1}}{{x}} \\ $$$${C}.\:\:{x}^{\mathrm{2}} \\ $$$${D}.\:\:{x} \\ $$ Commented…
Question Number 136761 by I want to learn more last updated on 25/Mar/21 Commented by I want to learn more last updated on 25/Mar/21 $$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}.…
Question Number 136757 by Dwaipayan Shikari last updated on 25/Mar/21 $$\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}} \\ $$ Answered by mindispower last…
Question Number 136760 by I want to learn more last updated on 25/Mar/21 Answered by mr W last updated on 25/Mar/21 $$\frac{{BC}}{\mathrm{sin}\:\angle{BDC}}=\frac{{CD}}{\mathrm{sin}\:\mathrm{30}°}\:\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\angle{BDA}}{{AB}}=\frac{\mathrm{sin}\:\mathrm{45}°}{{DA}}\:\:\:…\left({ii}\right) \\…
Question Number 136661 by Dwaipayan Shikari last updated on 24/Mar/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{16}}} {dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com