Question Number 164468 by ajfour last updated on 17/Jan/22 Answered by mr W last updated on 18/Jan/22 Commented by mr W last updated on 18/Jan/22…
Question Number 98675 by Algoritm last updated on 15/Jun/20 Commented by Algoritm last updated on 16/Jun/20 $$\mathrm{someone}\:\mathrm{gave}\:\mathrm{me}\:\mathrm{an}\:\mathrm{original}\:\mathrm{picture}\:\mathrm{and}\:\mathrm{I}'\mathrm{ll}\:\mathrm{translate}\:\mathrm{it} \\ $$ Commented by mr W last updated…
Question Number 164178 by mr W last updated on 15/Jan/22 Commented by mr W last updated on 15/Jan/22 $${two}\:{points}\:{P},\:{Q}\:{lie}\:{on}\:{the}\:{parabola} \\ $$$${with}\:{PQ}={b}.\:{find}\:{the}\:{locus}\:{of}\:{its} \\ $$$${midpoint}\:{M}.\:{find}\:{the}\:{lowest}\:{possible} \\ $$$${position}\:{of}\:{M}.…
Question Number 164174 by mr W last updated on 15/Jan/22 Commented by mr W last updated on 15/Jan/22 $${find}\:{the}\:{position}\:{of}\:{point}\:{P}\:{in}\:{a} \\ $$$${given}\:{triangle}\:{such}\:{that}\:{the}\:{sum} \\ $$$${of}\:{its}\:{distances}\:{to}\:{the}\:{vertices}\:{is} \\ $$$${minimum}.\:{find}\:{the}\:{corresponding}…
Question Number 98560 by I want to learn more last updated on 14/Jun/20 Commented by I want to learn more last updated on 14/Jun/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}.…
Question Number 32913 by 9588813186 last updated on 06/Apr/18 $$\mathrm{2}\wedge\mathrm{6} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163921 by ajfour last updated on 11/Jan/22 Answered by mr W last updated on 12/Jan/22 $${R}={radius}\:{of}\:{big}\:{circles} \\ $$$${r}={radius}\:{of}\:{small}\:{circles} \\ $$$${semi}\:{diagonal}={R}+\sqrt{\mathrm{2}}{R}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }+\sqrt{\mathrm{2}}{r} \\…
Question Number 32659 by Joel578 last updated on 30/Mar/18 Commented by Joel578 last updated on 30/Mar/18 $$\mathrm{Red}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cm} \\ $$$$\mathrm{Blue}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{4}\:\mathrm{cm} \\ $$$${C}\:\mathrm{is}\:\mathrm{center}\:\mathrm{of}\:\mathrm{blue}\:\mathrm{circle} \\ $$$${AB}\:\mathrm{is}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{red}\:\mathrm{circle} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part}…
Question Number 98190 by behi83417@gmail.com last updated on 12/Jun/20 Answered by mr W last updated on 12/Jun/20 Commented by mr W last updated on 12/Jun/20…
Question Number 163656 by mnjuly1970 last updated on 09/Jan/22 Answered by mr W last updated on 09/Jan/22 Commented by mr W last updated on 09/Jan/22…