Question Number 218812 by Spillover last updated on 15/Apr/25 Answered by A5T last updated on 16/Apr/25 $$\sqrt{\left(\mathrm{R}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{x}^{\mathrm{2}} −\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{R}−\mathrm{2x}\right)\mathrm{R}=\mathrm{h}\left(\mathrm{2x}−\mathrm{h}\right) \\ $$$$\Rightarrow\mathrm{R}^{\mathrm{2}}…
Question Number 218748 by MrGaster last updated on 15/Apr/25 $$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{cos}\:{x}}{\mathrm{cosh}\:\mathrm{2}{x}−\mathrm{cos}\:{x}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{e}^{\mathrm{4}{x}} −\mathrm{2}{e}^{\mathrm{2}{x}} \mathrm{cos}\:{x}+\mathrm{1}}{dx},\mathrm{lemma}:\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{cos}\:{kx}}{{p}^{{k}} }=\frac{\mathrm{p}\:\mathrm{cos}\:{x}−\mathrm{1}}{{p}^{\mathrm{2}} −\mathrm{2}{p}\:\mathrm{cos}\:{x}+\mathrm{1}},{p}>\mathrm{1} \\ $$ Answered by MrGaster…
Question Number 218813 by Spillover last updated on 15/Apr/25 Answered by nikif99 last updated on 15/Apr/25 Commented by Spillover last updated on 16/Apr/25 $${great}\:{work}.{thanks}\: \\…
Question Number 218749 by hardmath last updated on 15/Apr/25 Answered by MrGaster last updated on 17/Apr/25 $${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{ln}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}} }\right)}=\frac{\mathrm{1}}{\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}}…
Question Number 218799 by depressiveshrek last updated on 15/Apr/25 $$\mathrm{For}\:\mathrm{those}\:\mathrm{who}\:\mathrm{are}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{cryptography}. \\ $$$$\mathrm{The}\:\mathrm{below}\:\mathrm{text}\:\mathrm{has}\:\mathrm{been}\:\mathrm{encrypted}\:\mathrm{using} \\ $$$$\mathrm{Vigenere}\:\mathrm{cipher},\:\mathrm{such}\:\mathrm{that}\:\mathrm{numbers},\:\mathrm{punctuation} \\ $$$$\mathrm{marks}\:\mathrm{and}\:\mathrm{the}\:\mathrm{letter}\:\overset{..} {\mathrm{E}}\:\mathrm{have}\:\mathrm{remained}\:\mathrm{the}\:\mathrm{same}. \\ $$$$\mathrm{A}\:\mathrm{keyword}\:\mathrm{of}\:\mathrm{length}\:\mathrm{9}\:\mathrm{has}\:\mathrm{been}\:\mathrm{used},\:\mathrm{which} \\ $$$$\mathrm{starts}\:\mathrm{with}\:\mathrm{the}\:\mathrm{letter}\:\mathrm{K}.\:\mathrm{Decrypt}\:\mathrm{the}\:\mathrm{text}. \\ $$ Commented by…
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Question Number 218780 by Spillover last updated on 15/Apr/25 Commented by malwan last updated on 17/Apr/25 $${can}\:{you}\:{solve}\:{this}\:{integral} \\ $$$${please}\:? \\ $$ Answered by Spillover last…
Question Number 218781 by Ghisom last updated on 15/Apr/25 $$\mathrm{prove}: \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}}}\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$ Answered by breniam last updated on 20/Apr/25…
Question Number 218778 by Spillover last updated on 15/Apr/25 Answered by mr W last updated on 15/Apr/25 Commented by mr W last updated on 17/Apr/25…