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Question-228037

Question Number 228037 by fantastic2 last updated on 13/Mar/26 Answered by TonyCWX last updated on 14/Mar/26 $${b}\:=\:\mathrm{Base}\:\mathrm{of}\:\mathrm{triangle} \\ $$$${h}\:=\:\mathrm{Height}\:\mathrm{of}\:\mathrm{triangle} \\ $$$${h}\:>\:{b} \\ $$$$ \\ $$$${bh}\:=\:\mathrm{12}…

Question-228034

Question Number 228034 by Zouhir last updated on 13/Mar/26 Answered by Ghisom_ last updated on 13/Mar/26 $$\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{2sin}\:{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right) \\ $$$$\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\int{dx}−\int\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:{dx}\right)= \\ $$$$=\frac{{x}}{\mathrm{2}}+\mathrm{ln}\:\mid\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\mid\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\pi}{\mathrm{4}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}…

Question-228029

Question Number 228029 by Kassista last updated on 11/Mar/26 Commented by Ghisom_ last updated on 12/Mar/26 $$\mathrm{reminds}\:\mathrm{me}\:\mathrm{of}\:\mathrm{the}\:\mathrm{once}\:\mathrm{famous}\:\mathrm{story}\:\mathrm{by} \\ $$$$\mathrm{Jorge}\:\mathrm{Luis}\:\mathrm{Borges}\:“\mathrm{The}\:\mathrm{Library}\:\mathrm{of}\:\mathrm{Babel}'' \\ $$$$ \\ $$$$\mathrm{everybody}\:\mathrm{who}\:\mathrm{deals}\:\mathrm{with}\:\mathrm{any}\:\mathrm{kind}\:\mathrm{of} \\ $$$$\mathrm{infinities}\:\mathrm{should}\:\mathrm{read}\:\mathrm{it}……

Question-227916

Question Number 227916 by Lara2440 last updated on 01/Mar/26 Answered by Kassista last updated on 01/Mar/26 $$ \\ $$$${let}\:{A}=\left[\mathrm{0},\mathrm{1}\right]\:{and}\:{B}=\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} \\ $$$${there}\:{is}\:{an}\:{injection}\:{from}\:{A}\:{to}\:{B} \\ $$$${f}:{A}\rightarrow{B}\:{s}.{t}.\:{f}\left({x}\right)=\left({x},\mathrm{0}\right) \\ $$$${pf}:\:{given}\:{a},{b}\in{A}…

Question-227909

Question Number 227909 by Chiblinkz last updated on 27/Feb/26 Answered by gregori last updated on 28/Feb/26 $$\left({c}\right)\:\lambda^{\mathrm{2}} −\lambda−\mathrm{1}=\mathrm{0}\:\Rightarrow\:\lambda=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:{U}_{{n}} =\:{p}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{q}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$$\Rightarrow{p}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+{q}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\:\mathrm{1} \\…

Set-Theory-amp-Analysis-I-m-stuck-on-a-set-theory-problem-in-my-analysis-textbook-it-asks-to-prove-the-existence-of-a-bijection-function-between-then-unit-interval-0-1-and-the-0-1-0-1-Hone

Question Number 227905 by Lara2440 last updated on 28/Feb/26 $$\: \\ $$$$\:\mathrm{Set}\:\mathrm{Theory\&Analysis} \\ $$$$\: \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{stuck}\:\mathrm{on}\:\mathrm{a}\:\mathrm{set}\:\mathrm{theory}\:\mathrm{problem}\:\mathrm{in}\:\mathrm{my}\:\mathrm{analysis}\:\mathrm{textbook}. \\ $$$$\mathrm{it}\:\mathrm{asks}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{existence}\:\mathrm{of}\:\mathrm{a}\:\mathrm{bijection}\:\mathrm{function} \\ $$$$\mathrm{between}\:\mathrm{then}\:\mathrm{unit}\:\mathrm{interval}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{and}\:\mathrm{the}\:\left[\mathrm{0},\mathrm{1}\right]×\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{Honestly}\:\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{clue}\:\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{this}…. \\ $$$$\mathrm{So}\:\mathrm{I}\:\mathrm{just}\:\mathrm{skipped}\:\mathrm{it}\:\mathrm{for}\:\mathrm{now}. \\…

Question-Suppose-that-we-consider-the-expression-1-1-n-n-as-n-the-sequence-converges-to-a-unique-irrational-constant-denoted-by-e-def-lim-n-1-1-n-n-e-Let-Sequence-A-h-1-1

Question Number 227865 by Lara2440 last updated on 23/Feb/26 $$\mathrm{Question}. \\ $$$$\: \\ $$$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{we}\:\mathrm{consider}\:\mathrm{the}\:\mathrm{expression}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} .\: \\ $$$$\mathrm{as}\:{n}\rightarrow\infty\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{converges}\:\mathrm{to}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{irrational}\:\mathrm{constant}\:\mathrm{denoted}\:\mathrm{by}\:''\boldsymbol{\mathrm{e}}'' \\ $$$$\mathrm{def}.\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} ={e} \\ $$$$\mathrm{Let}\:\mathrm{Sequence}\:{A}_{{h}}…