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Author: Tinku Tara

2021-HAPPY-NEW-Year-1-x-3-3x-2-x-2-1-2-x-1-dx-2-2cos-x-sin-x-3sin-x-5cos-x-dx-3-tan-2x-sin-6-x-cos-6-x-dx-4-x-1-x-2-1-x-2-dx-

Question Number 127870 by Eric002 last updated on 02/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2021} \\ $$$${HAPPY}\:{NEW}\:{Year} \\ $$$$\left.\mathrm{1}\right)\int\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{dx} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\int\frac{\mathrm{2}{cos}\left({x}\right)−{sin}\left({x}\right)}{\mathrm{3}{sin}\left({x}\right)+\mathrm{5}{cos}\left({x}\right)}{dx} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\int\frac{{tan}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}^{\mathrm{6}}…

1-calculate-f-x-y-0-e-xt-cos-yt-t-dt-and-g-x-y-0-e-xt-sin-yt-t-dt-with-x-gt-0-and-y-gt-0-2-find-the-values-of-0-e-2t-cos-t-t-dt-and-0-

Question Number 62335 by maxmathsup by imad last updated on 19/Jun/19 $$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x},{y}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{xt}} {cos}\left({yt}\right)}{\:\sqrt{{t}}}\:{dt}\:{and}\:{g}\left({x},{y}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{xt}} {sin}\left({yt}\right)}{\:\sqrt{{t}}}\:{dt} \\ $$$${with}\:{x}>\mathrm{0}\:\:{and}\:{y}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−\mathrm{2}{t}} \:{cos}\left({t}\right)}{\:\sqrt{{t}}}\:{dt}\:{and}\:\int_{\mathrm{0}}…

if-2-2-2-2-evaluate-

Question Number 62334 by smartsmith459@gmail.com last updated on 19/Jun/19 $${if}\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\:{evaluate}\left(\alpha−\beta\right) \\ $$ Answered by Kunal12588 last updated on 20/Jun/19 $$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}}…

Question-62332

Question Number 62332 by tanmay last updated on 19/Jun/19 Answered by tanmay last updated on 20/Jun/19 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\sqrt{\mathrm{2}}\:×{n}^{{n}−\frac{\mathrm{1}}{\mathrm{2}}} }{{n}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} ×\sqrt{\mathrm{2}\pi}\:×{e}^{−{n}} }×\left\{\frac{\left(\mathrm{2}×{n}^{\frac{\mathrm{1}}{{n}}} −\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\right\}^{\frac{{n}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{{ln}^{\mathrm{2}} {n}}}…

Question-127865

Question Number 127865 by Ar Brandon last updated on 02/Jan/21 Answered by mathmax by abdo last updated on 03/Jan/21 $$\left.\mathrm{a}\right)\mathrm{wehave}\:\mathrm{u}_{\mathrm{n}+\mathrm{1}\:} =\mathrm{f}\left(\mathrm{u}_{\mathrm{n}} \right)\:\mathrm{with}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}−\mathrm{x}^{\mathrm{2}} \:\:\mathrm{f}\:\mathrm{is}\:\mathrm{contnue}\: \\ $$$$\mathrm{the}\:\mathrm{fix}\:\mathrm{point}\:\:\mathrm{verify}\:\mathrm{x}=\mathrm{x}−\mathrm{x}^{\mathrm{2}}…

How-many-natural-numbers-n-2020-such-that-n-3-is-divisible-by-2-n-

Question Number 127861 by naka3546 last updated on 02/Jan/21 $${How}\:\:{many}\:\:{natural}\:\:{numbers}\:\:{n}\:\leqslant\:\mathrm{2020}\:{such}\:\:{that} \\ $$$$\left({n}\:+\:\mathrm{3}\right)!\:\:{is}\:\:{divisible}\:\:{by}\:\:\mathrm{2}^{{n}} \:? \\ $$ Answered by floor(10²Eta[1]) last updated on 03/Jan/21 $$\mathrm{2}^{\mathrm{n}} \mid\left(\mathrm{n}+\mathrm{3}\right)!\Rightarrow\mathrm{k}\geqslant\mathrm{n} \\…

Question-127859

Question Number 127859 by AgnibhoMukhopadhyay last updated on 02/Jan/21 Answered by liberty last updated on 03/Jan/21 $$\:\frac{\mathrm{3}−\mathrm{5i}}{\mathrm{8}+\mathrm{2i}}\:×\:\frac{\mathrm{8}−\mathrm{2i}}{\mathrm{8}−\mathrm{2i}}\:=\:\frac{\mathrm{24}−\mathrm{46i}−\mathrm{10}}{\mathrm{64}+\mathrm{4}}=\frac{\mathrm{14}−\mathrm{46i}}{\mathrm{68}} \\ $$$$\:=\:\frac{\mathrm{7}}{\mathrm{34}}−\frac{\mathrm{23i}}{\mathrm{34}} \\ $$ Terms of Service Privacy…

Question-62322

Question Number 62322 by aliesam last updated on 19/Jun/19 Commented by arcana last updated on 19/Jun/19 $$\beta_{\mathrm{1}} =\alpha+\alpha^{\mathrm{6}} \\ $$$$\beta_{\mathrm{2}} =\left(\alpha+\alpha^{\mathrm{6}} \right)^{\mathrm{2}} −\mathrm{2}=\alpha^{\mathrm{2}} +\mathrm{2}\alpha^{\mathrm{7}} +\alpha^{\mathrm{12}}…