Question Number 219067 by Spillover last updated on 19/Apr/25 Commented by Spillover last updated on 19/Apr/25 Quarter circle area = Semicircle area. Green area/Orange area = ? Answered by mr W last updated on 20/Apr/25…
Question Number 219060 by MrGaster last updated on 19/Apr/25 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{{m}} }{{x}^{{n}} }{dx},{n}\in\mathbb{N},{m}\in\mathbb{N},{n}\leqslant{m} \\ $$ Answered by Nicholas666 last updated on 19/Apr/25 $$\frac{\pi}{\mathrm{2}} \\…
Question Number 219120 by Spillover last updated on 19/Apr/25 Answered by mr W last updated on 20/Apr/25 Commented by mr W last updated on 20/Apr/25…
Question Number 219116 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $$\mathrm{tan30}°=\frac{\mathrm{r}}{\mathrm{x}}\Rightarrow\mathrm{x}=\mathrm{r}\sqrt{\mathrm{3}} \\ $$$$\mathrm{k}=\mathrm{sin15}°=\frac{\mathrm{R}}{\left(\mathrm{2r}+\mathrm{2r}\sqrt{\mathrm{3}}\right)−\mathrm{R}} \\ $$$$\Rightarrow\mathrm{R}+\mathrm{Rk}=\mathrm{kr}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\right)\Rightarrow\frac{\mathrm{r}}{\mathrm{R}}=\frac{\mathrm{1}+\mathrm{k}}{\mathrm{k}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\right)} \\ $$$$\mathrm{sin30}°=\mathrm{2sin15}°\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{15}°} \\…
Question Number 219117 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $$\left(\mathrm{r}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{r}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4r}=\mathrm{6}+\mathrm{r}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{r}=\frac{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}−\sqrt{\mathrm{3}}} \\ $$…
Question Number 219118 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $$\mathrm{R}^{\mathrm{2}} =\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \Rightarrow\mathrm{R}=\frac{\mathrm{a}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{L}=\mathrm{R}−\frac{\mathrm{a}}{\mathrm{2}}=\frac{\mathrm{a}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{a}}{\mathrm{L}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\…
Question Number 219119 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $$\mathrm{R}=\mathrm{6} \\ $$$$\mathrm{Area}=\mathrm{4}×\left[\mathrm{2}\left(\frac{\mathrm{6}^{\mathrm{2}} \pi}{\mathrm{4}}−\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}\right)\right]=\mathrm{8}\left[\frac{\mathrm{36}\pi−\mathrm{72}}{\mathrm{4}}\right]=\mathrm{72}\left(\pi−\mathrm{2}\right) \\ $$ Answered by…
Question Number 219055 by lengad last updated on 19/Apr/25 Commented by mr W last updated on 19/Apr/25 $${please}\:{post}\:{clear}\:{and}\:{properly}\: \\ $$$${cropped}\:{picture}\centerdot \\ $$ Terms of Service…
Question Number 219112 by hardmath last updated on 19/Apr/25 $$\mathrm{Prove}\:\mathrm{it}: \\ $$$$\mathrm{In}\:\mathrm{triangle}\:\mathrm{ABC},\:\mathrm{AB}=\mathrm{c},\:\mathrm{BC}=\mathrm{b},\:\mathrm{AC}=\mathrm{a} \\ $$$$\mathrm{ab}^{\mathrm{2}} \mathrm{c}\:+\:\mathrm{abc}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \mathrm{bc}\:\geqslant\:\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}\:\frac{\mathrm{2S}^{\mathrm{3}} }{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} } \\ $$ Terms of Service…
Question Number 219113 by hardmath last updated on 19/Apr/25 $$\frac{\mathrm{a}\:+\:\mathrm{3b}}{\mathrm{a}\:+\:\mathrm{b}−\mathrm{1}}\:+\:\frac{\mathrm{a}\:+\:\mathrm{3b}−\mathrm{1}}{\mathrm{a}\:+\:\mathrm{b}−\mathrm{3}}\:=\:\mathrm{4}\:\:\Rightarrow\:\:\mathrm{a}\:+\:\mathrm{b}\:=\:? \\ $$ Commented by Rasheed.Sindhi last updated on 19/Apr/25 $$\mathrm{a}+\mathrm{b}\:\mathrm{is}\:\mathrm{not}\:\mathrm{unique},\: \\ $$ Answered by Rasheed.Sindhi…