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1-1-3-2-1-3-5-3-1-3-5-7-n-terms-Find-sum-

Question Number 105250 by bshahid010@gmail.com last updated on 27/Jul/20 $$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}×\mathrm{3}×\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}}+………\mathrm{n}−\mathrm{terms} \\ $$$$\mathrm{Find}\:\mathrm{sum} \\ $$ Answered by nimnim last updated on 27/Jul/20 $${Let}\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}×\mathrm{3}×\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}}+….{to}\:{n}\:{terms} \\ $$$${Nr}={n}^{{th}}…

lim-x-0-cosx-1-x-2-

Question Number 105246 by moonanddolldoll last updated on 27/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}{x}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$ Answered by bemath last updated on 27/Jul/20 $${another}\:{method} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\left(\mathrm{cos}\:{x}−\mathrm{1}\right)\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}}…

Solve-x-y-3-i-x-y-5-ii-

Question Number 170754 by Tawa11 last updated on 30/May/22 $$\mathrm{Solve}: \\ $$$$\:\:\:\:\:\mathrm{x}\:\:\:\:+\:\:\:\sqrt{\mathrm{y}}\:\:\:\:=\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:……\:\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\sqrt{\mathrm{x}}\:\:\:\:+\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:……\:\:\left(\mathrm{ii}\right) \\ $$ Commented by MJS_new last updated on 30/May/22 $$\mathrm{obviously}\:{x}=\mathrm{1}\wedge{y}=\mathrm{4} \\…

3-sin-2-40-1-cos-2-40-64-sin-2-40-

Question Number 170737 by naka3546 last updated on 30/May/22 $$\frac{\mathrm{3}}{\mathrm{sin}^{\mathrm{2}} \mathrm{40}°}\:−\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{40}°}\:+\:\mathrm{64}\:\mathrm{sin}^{\mathrm{2}} \mathrm{40}°\:\:=\:\:? \\ $$ Answered by som(math1967) last updated on 30/May/22 $$\frac{\mathrm{3}}{{sin}^{\mathrm{2}} \mathrm{40}}\:+\frac{\mathrm{64}{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}}…

Question-105195

Question Number 105195 by mohammad17 last updated on 26/Jul/20 Answered by Aziztisffola last updated on 26/Jul/20 $$\mathrm{L}\left(\mathrm{2y}''+\mathrm{3y}'−\mathrm{2y}\right)=\mathrm{L}\left(\mathrm{te}^{−\mathrm{2t}} \right) \\ $$$$\mathrm{2s}^{\mathrm{2}} \mathrm{Y}\left(\mathrm{s}\right)−\mathrm{2sy}\left(\mathrm{0}\right)−\mathrm{2y}\left(\mathrm{0}\right)+\mathrm{3sY}\left(\mathrm{s}\right)−\mathrm{3y}\left(\mathrm{0}\right)−\mathrm{2Y}\left(\mathrm{s}\right)=−\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathrm{L}\left(\mathrm{e}^{−\mathrm{2t}} \right)\right) \\ $$$$\left(\mathrm{2s}^{\mathrm{2}} +\mathrm{3s}−\mathrm{2}\right)\mathrm{Y}\left(\mathrm{s}\right)+\mathrm{4}=−\frac{\mathrm{d}}{\mathrm{ds}}\left(\frac{\mathrm{1}}{\mathrm{s}+\mathrm{2}}\right)…

Question-39651

Question Number 39651 by KMA last updated on 09/Jul/18 $$ \\ $$ Answered by MJS last updated on 09/Jul/18 $$−\frac{\mathrm{2}}{\mathrm{25e}^{\sqrt{{x}}} }\left(\mathrm{2}\left(\mathrm{5}\sqrt{{x}}+\mathrm{2}\right)\mathrm{cos}\left(\mathrm{2}\sqrt{{x}}−\mathrm{3}\right)+\left(\sqrt{\mathrm{5}}{x}−\mathrm{3}\right)\mathrm{sin}\left(\mathrm{2}\sqrt{{x}}−\mathrm{3}\right)\right)+{C} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{question}\:\mathrm{LOL} \\ $$…