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Author: Tinku Tara

14-1-x-ln-x-3-d-x-

Question Number 203061 by York12 last updated on 08/Jan/24 $$\mathrm{14}\int\frac{\mathrm{1}}{\mathrm{x}\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{3}} \:}\:\:\mathrm{d}\left(\mathrm{x}\right) \\ $$ Answered by MM42 last updated on 09/Jan/24 $${ans} \\ $$$$\:\frac{−\mathrm{7}}{\left({lnx}\right)^{\mathrm{2}} } \\…

Question-203059

Question Number 203059 by hassanmpsy last updated on 08/Jan/24 Commented by witcher3 last updated on 11/Jan/24 $$\mathrm{U}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{n}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)}{\mathrm{n}^{\mathrm{2}} \left(\mathrm{2}+\mathrm{2}\frac{\mathrm{k}}{\mathrm{n}}+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{f}\left(\frac{\mathrm{k}}{\mathrm{n}}\right) \\…

A-two-digit-number-AB-10-AB-10-A-B-BA-10-Find-the-number-There-is-a-poem-Having-arrived-at-the-age-of-BA-10-

Question Number 203051 by ajfour last updated on 08/Jan/24 $${A}\:{two}\:{digit}\:{number}\:\:\left({AB}\right)_{\mathrm{10}} \\ $$$$\left({AB}\right)_{\mathrm{10}} −{A}^{{B}} =\left({BA}\right)_{\mathrm{10}} \\ $$$${Find}\:{the}\:{number}.\:{There}\:{is}\:{a}\:{poem}. \\ $$$${Having}\:{arrived}\:{at}\:{the}\:{age}\:{of}\:\left({BA}\right)_{\mathrm{10}} . \\ $$ Answered by Rasheed.Sindhi last…

e-2-Proof-Let-x-e-2-2-2x-e-2-2x-e-2-e-2-e-2-2ex-4x-e-2-4-4x-4-2ex-e-2-x-2-4x-4-x-2-2ex-e-2-x-2-2-x-e-2-x-2-2-x-e-2-x-2-x-e-2-e-e-2-

Question Number 203035 by Frix last updated on 07/Jan/24 $$\mathrm{e}=\mathrm{2} \\ $$$$\mathrm{Proof}: \\ $$$$\mathrm{Let}\:{x}=\frac{\mathrm{e}+\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{2}{x}=\mathrm{e}+\mathrm{2} \\ $$$$\mathrm{2}{x}\left(\mathrm{e}−\mathrm{2}\right)=\left(\mathrm{e}+\mathrm{2}\right)\left(\mathrm{e}−\mathrm{2}\right) \\ $$$$\mathrm{2e}{x}−\mathrm{4}{x}=\mathrm{e}^{\mathrm{2}} −\mathrm{4} \\ $$$$−\mathrm{4}{x}+\mathrm{4}=−\mathrm{2e}{x}+\mathrm{e}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}}…