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Question-225892

Question Number 225892 by mnjuly1970 last updated on 15/Nov/25 Answered by A5T last updated on 15/Nov/25 $$\mathrm{Let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{and}\:\mathrm{OX}=\mathrm{d} \\ $$$$\mathrm{Let}\:\mathrm{OB}=\mathrm{r}\Rightarrow\mathrm{XB}=\mathrm{r}−\mathrm{d};\:\mathrm{AX}=\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{e} \\ $$$$\mathrm{AO}=\sqrt{\mathrm{e}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} }\: \\ $$$$\angle\mathrm{BYC}=\mathrm{90}°\:\Rightarrow\:\mathrm{XHYC}\:\mathrm{is}\:\mathrm{cyclic}…

0-1-ln-12-1-x-ln-12-1-x-ln-12-1-x-ln-12-1-x-dx-

Question Number 225861 by fantastic2 last updated on 15/Nov/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{{ln}^{\mathrm{12}} \left(\mathrm{1}−{x}\right)}{\:\sqrt{\frac{{ln}^{\mathrm{12}} \left(\mathrm{1}−{x}\right)}{\:\sqrt{\frac{{ln}^{\mathrm{12}} \left(\mathrm{1}−{x}\right)}{\:\sqrt{\frac{{ln}^{\mathrm{12}} \left(\mathrm{1}−{x}\right)}{…}}}}}}}}{dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-225856

Question Number 225856 by ajfour last updated on 15/Nov/25 Commented by ajfour last updated on 15/Nov/25 $${Parabola}\:{shown}\:{is}\:\:{y}={x}^{\mathrm{2}} . \\ $$$${Find}\:{equations}\:{of}\:{maximum} \\ $$$${radius}\:{yellow}\:{circles},\:{or}\:{find} \\ $$$$\:\:\:\:\:\:\:{r},\:{C}\left({h},{k}\right)\:,\:{C}\:'\left(−{h},{k}\right) \\…

Prove-that-in-any-triangle-4R-r-w-a-w-b-w-c-h-a-h-b-h-c-1-a-1-b-a-b-2-

Question Number 225810 by hardmath last updated on 12/Nov/25 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{any}\:\mathrm{triangle}: \\ $$$$\frac{\mathrm{4R}}{\mathrm{r}}\:\geqslant\:\frac{\mathrm{w}_{\boldsymbol{\mathrm{a}}} \:\mathrm{w}_{\boldsymbol{\mathrm{b}}} \:\mathrm{w}_{\boldsymbol{\mathrm{c}}} }{\mathrm{h}_{\boldsymbol{\mathrm{a}}} \:\mathrm{h}_{\boldsymbol{\mathrm{b}}} \:\mathrm{h}_{\boldsymbol{\mathrm{c}}} }\:\centerdot\:\left(\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\right)\centerdot\left(\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\right)^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy…