Question Number 43023 by ajfour last updated on 06/Sep/18 Commented by ajfour last updated on 06/Sep/18 $${Find}\:{location}\:{of}\:{centre}\left(\mathrm{0},{h}\right)\:{and} \\ $$$${radius}\:{R}\:{of}\:{circle}\:{such}\:{that} \\ $$$${all}\:{four}\:{bounded}\:{areas}\: \\ $$$${O}\overset{\frown} {{AD}}\:,\:\overset{\frown} {{APB}A}\:,\:{A}\overset{\frown}…
Question Number 108461 by ajfour last updated on 17/Aug/20 Commented by ajfour last updated on 17/Aug/20 $${If}\:{the}\:{two}\:{shaded}\:{areas}\:{are}\:{equal}, \\ $$$${find}\:\alpha\:{in}\:{terms}\:{of}\:{radius} \\ $$$${ratio}\:\:{a}/{b}. \\ $$ Answered by…
Question Number 42866 by Cheyboy last updated on 03/Sep/18 Answered by MJS last updated on 04/Sep/18 $${x}\rightarrow{r}\:\mathrm{to}\:\mathrm{not}\:\mathrm{get}\:\mathrm{confused} \\ $$$$\mathrm{small}\:\mathrm{circle}\:\mathrm{touching}\:\:\mathrm{half}\:\mathrm{circle}\:\mathrm{from}\:\mathrm{inside} \\ $$$${r}^{\mathrm{2}} +{p}^{\mathrm{2}} −\left(\frac{{R}}{\mathrm{2}}−{r}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{p}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}}…
Question Number 173870 by savitar last updated on 20/Jul/22 $$\sqrt{\mathrm{1}+\:\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\left({n}+\mathrm{4}\right)}\:\in\:\mathbb{N} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 173124 by peter frank last updated on 06/Jul/22 Commented by aleks041103 last updated on 06/Jul/22 $${For}\:{which}…{what}? \\ $$ Commented by peter frank last…
Question Number 41252 by ajfour last updated on 04/Aug/18 Commented by ajfour last updated on 04/Aug/18 $${What}\:{fraction}\:{of}\:{volume}\:{of}\:{cup} \\ $$$${is}\:{filled}\:? \\ $$ Answered by MrW3 last…
Question Number 172269 by Mikenice last updated on 25/Jun/22 $${A}\:{rectangular}\:{picture}\:\mathrm{6}{cm}\:{by}\:\mathrm{8}{cm}\:{is} \\ $$$${enclosed}\:{by}\:{a}\:{frame}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:{wide}.\: \\ $$$${calculate}\:{the}\left[{area}\:{of}\:{the}\:{frame}.\right. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 172266 by Mikenice last updated on 25/Jun/22 Answered by mr W last updated on 25/Jun/22 Commented by mr W last updated on 25/Jun/22…
Question Number 41103 by ajfour last updated on 02/Aug/18 Commented by ajfour last updated on 02/Aug/18 $${Q}.\mathrm{41062}\:\:{solution} \\ $$ Commented by MJS last updated on…
Question Number 105883 by bemath last updated on 01/Aug/20 Answered by mr W last updated on 01/Aug/20 $${AG}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{2}×\mathrm{3}×\mathrm{6}×\mathrm{cos}\:\mathrm{120}°}=\mathrm{3}\sqrt{\mathrm{7}} \\ $$$$\frac{\mathrm{sin}\:\angle{AGF}}{\mathrm{3}}=\frac{\mathrm{sin}\:\mathrm{120}°}{\mathrm{3}\sqrt{\mathrm{7}}} \\ $$$$\mathrm{sin}\:\angle{AGF}=\frac{\sqrt{\mathrm{21}}}{\mathrm{14}} \\…