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Question Number 228047 by mr W last updated on 14/Mar/26 Commented by mr W last updated on 14/Mar/26 $${find}\:{the}\:{area}\:{of}\:{the}\:{red}\:{circle}. \\ $$ Answered by TonyCWX last…
Question Number 228049 by Spillover last updated on 14/Mar/26 Answered by Kassista last updated on 14/Mar/26 $${in}\:{caseB},\:{part}\:{of}\:{the}\:{force},{F}\mathrm{sin}\:\left(\theta\right),\:{is}\:{being}\:{used}\:{upward} \\ $$$${which}\:{is}\:{useless},\:{thefore},\:{Person}\:{A}\:{is}\:{smarter} \\ $$ Commented by fantastic2 last…
Question Number 228037 by fantastic2 last updated on 13/Mar/26 Answered by TonyCWX last updated on 14/Mar/26 $${b}\:=\:\mathrm{Base}\:\mathrm{of}\:\mathrm{triangle} \\ $$$${h}\:=\:\mathrm{Height}\:\mathrm{of}\:\mathrm{triangle} \\ $$$${h}\:>\:{b} \\ $$$$ \\ $$$${bh}\:=\:\mathrm{12}…
Question Number 228034 by Zouhir last updated on 13/Mar/26 Answered by Ghisom_ last updated on 13/Mar/26 $$\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) \\ $$$$\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\int{dx}−\int\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:{dx}\right)= \\ $$$$=\frac{{x}}{\mathrm{2}}+\mathrm{ln}\:\mid\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\mid\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\pi}{\mathrm{4}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}…
Question Number 228030 by TonyCWX last updated on 12/Mar/26 $${I}\:=\:\int\:\frac{\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{1}}\:{dx}\:=\:\int\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\:{dx} \\ $$$$=\:\int\:\left[\frac{\mathrm{1}}{\mathrm{5}\left({x}+\mathrm{1}\right)}\:−\:\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{\mathrm{5}\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\right]\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}}\int\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:{dx}\:−\:\frac{\mathrm{1}}{\mathrm{5}}\int\:\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{{x}^{\mathrm{4}}…
Question Number 228029 by Kassista last updated on 11/Mar/26 Commented by Ghisom_ last updated on 12/Mar/26 $$\mathrm{reminds}\:\mathrm{me}\:\mathrm{of}\:\mathrm{the}\:\mathrm{once}\:\mathrm{famous}\:\mathrm{story}\:\mathrm{by} \\ $$$$\mathrm{Jorge}\:\mathrm{Luis}\:\mathrm{Borges}\:“\mathrm{The}\:\mathrm{Library}\:\mathrm{of}\:\mathrm{Babel}'' \\ $$$$ \\ $$$$\mathrm{everybody}\:\mathrm{who}\:\mathrm{deals}\:\mathrm{with}\:\mathrm{any}\:\mathrm{kind}\:\mathrm{of} \\ $$$$\mathrm{infinities}\:\mathrm{should}\:\mathrm{read}\:\mathrm{it}……
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Question Number 228019 by mr W last updated on 10/Mar/26 $${t}\:{is}\:{the}\:{fractional}\:{part}\:{of}\:{a},\:{and} \\ $$$${a}^{\mathrm{2}} +{t}^{\mathrm{2}} =\mathrm{18}.\:{find}\:{t}=? \\ $$ Answered by Ghisom_ last updated on 10/Mar/26 $$\forall{r}\in\mathbb{R}:\:\mathrm{fpart}\:\left({r}\right)\:={r}−\mathrm{ipart}\:\left({r}\right)…
Question Number 228013 by Mingma last updated on 09/Mar/26 Answered by A5T last updated on 14/Mar/26 $$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{AB}=\mathrm{1}\:\mathrm{and}\:\mathrm{BC}=\mathrm{x}\:;\:\mathrm{then}\:\mathrm{AC}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\bigtriangleup\mathrm{CBD}\approx\bigtriangleup\mathrm{CAB}\Rightarrow\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{DC}}{\mathrm{BC}}\:\Rightarrow\mathrm{DC}=\frac{\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{BD}=\sqrt{\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{x}^{\mathrm{4}}…