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Author: Tinku Tara

nice-calculus-prove-that-0-1-ln-x-1-x-2-x-dx-pi-2-16-

Question Number 137439 by mnjuly1970 last updated on 02/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:{calculus}….. \\ $$$$\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\chi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:…. \\ $$ Answered by mindispower last updated…

A-nuclide-81-210-X-decays-to-another-nuclide-80-A-Y-in-four-successive-radioactive-decays-Each-decay-involves-the-emmision-of-either-an-alpha-particle-or-a-beta-particle-The-value-of-A-is-

Question Number 137429 by physicstutes last updated on 02/Apr/21 $$\mathrm{A}\:\mathrm{nuclide}\:_{\mathrm{81}} ^{\mathrm{210}} {X}\:\mathrm{decays}\:\mathrm{to}\:\mathrm{another}\:\mathrm{nuclide}\:_{\mathrm{80}} ^{{A}} {Y}\:\mathrm{in}\: \\ $$$$\mathrm{four}\:\mathrm{successive}\:\mathrm{radioactive}\:\mathrm{decays}.\:\mathrm{Each}\:\mathrm{decay} \\ $$$$\mathrm{involves}\:\mathrm{the}\:\mathrm{emmision}\:\mathrm{of}\:\mathrm{either}\:\mathrm{an}\:\mathrm{alpha}\:\mathrm{particle} \\ $$$$\mathrm{or}\:\mathrm{a}\:\mathrm{beta}\:\mathrm{particle}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{A}\:\mathrm{is}: \\ $$$$\mathrm{A}.\:\mathrm{120}\:\:\:\:\:\:\:\:\:\:\:\mathrm{B}.\:\mathrm{206} \\ $$$$\mathrm{C}.\:\mathrm{208}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}.\:\mathrm{212} \\…

lim-x-pi-2-sin-x-tanx-

Question Number 71894 by 20190927 last updated on 21/Oct/19 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{tanx}} \\ $$ Commented by kaivan.ahmadi last updated on 21/Oct/19 $$={e}^{{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \left({sinx}−\mathrm{1}\right){tanx}} ={e}^{{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \frac{{sinx}−\mathrm{1}}{{cotx}}} =\:\:…

i-n-1-3i-1-2-3-n-1-i-0-

Question Number 6354 by sanusihammed last updated on 24/Jun/16 $${i}\:=\:{n}−\mathrm{1} \\ $$$$\Sigma\:\:\mathrm{3}{i}\:\:\:\:\:=\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}^{{n}} \:−\:\mathrm{1}\right) \\ $$$${i}\:=\:\mathrm{0} \\ $$ Commented by nburiburu last updated on 24/Jun/16 $${doesn}'{t}\:{seem}\:{well}\:{writen}:…

1-2-3-2-5-2-2n-1-2-n-3-4n-2-1-

Question Number 6353 by sanusihammed last updated on 24/Jun/16 $$\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:\mathrm{5}^{\mathrm{2}} \:+\:………\:+\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\frac{{n}}{\mathrm{3}}\left(\mathrm{4}{n}^{\mathrm{2}\:} −\:\mathrm{1}\right) \\ $$ Commented by prakash jain last updated on 24/Jun/16…