Question Number 43728 by Cheyboy last updated on 14/Sep/18 Commented by A.Haq.Soomro last updated on 15/Sep/18 $$\mathrm{See}\:\mathrm{Q}#\mathrm{6852} \\ $$ Answered by MrW3 last updated on…
Question Number 43531 by pieroo last updated on 11/Sep/18 $$\mathrm{the}\:\mathrm{circumference}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{track}\:\mathrm{is}\:\mathrm{9}\:\mathrm{km}.\:\mathrm{A}\:\mathrm{cyclist} \\ $$$$\mathrm{rides}\:\mathrm{round}\:\mathrm{it}\:\mathrm{a}\:\mathrm{number}\:\mathrm{of}\:\mathrm{times}\:\mathrm{and}\:\mathrm{stops}\:\mathrm{after} \\ $$$$\mathrm{covering}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{302}\:\mathrm{km}.\:\mathrm{How}\:\mathrm{far}\:\mathrm{is}\:\mathrm{the}\:\mathrm{cyclist} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{point}?\:\left[\mathrm{take}\:\Pi=\frac{\mathrm{22}}{\mathrm{7}}\right] \\ $$ Commented by pieroo last updated on 12/Sep/18…
Question Number 108987 by bobhans last updated on 20/Aug/20 Answered by bemath last updated on 20/Aug/20 $$\:\:\:\:\frac{\mathcal{B}{e}\mathcal{M}{ath}}{\blacksquare\mathcal{C}\square\square\mathcal{L}\blacksquare} \\ $$$${because}\:\mid{x}−\mathrm{3}\mid\:+\:\mathrm{2}\:>\:\mathrm{0}\:{for}\:{x}\:\in\:\mathbb{R} \\ $$$${then}\:\mathrm{3}−\mid{x}−\mathrm{3}\mid\:<\:\mathrm{4}\mid{x}−\mathrm{3}\mid\:+\:\mathrm{8} \\ $$$$\Rightarrow−\mathrm{5}\mid{x}−\mathrm{3}\mid\:<\:\mathrm{5}\: \\ $$$$\Rightarrow\mid{x}−\mathrm{3}\mid\:>\:−\mathrm{2}\:\Rightarrow\:\forall{x}\in\mathbb{R}…
Question Number 43252 by ajfour last updated on 08/Sep/18 Commented by MJS last updated on 08/Sep/18 $$\mathrm{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${y}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{parabola}\:{x}={ay}^{\mathrm{2}} −\mathrm{1}…
Question Number 43131 by ajfour last updated on 07/Sep/18 Commented by ajfour last updated on 07/Sep/18 $${Q}.\mathrm{43128}\:\:\left({A}\:{possible}\:{solution}\right) \\ $$ Answered by ajfour last updated on…
Question Number 43128 by Cheyboy last updated on 07/Sep/18 Commented by Cheyboy last updated on 07/Sep/18 $${Find}\:{the}\:{area} \\ $$ Commented by ajfour last updated on…
Question Number 108643 by ajfour last updated on 18/Aug/20 Answered by mr W last updated on 18/Aug/20 Commented by mr W last updated on 18/Aug/20…
Question Number 43023 by ajfour last updated on 06/Sep/18 Commented by ajfour last updated on 06/Sep/18 $${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}}…