Question Number 92116 by M±th+et+s last updated on 04/May/20 $${what}\:{is}\:{the}\:{super}\:{hexagon}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 91836 by M±th+et+s last updated on 03/May/20 Commented by M±th+et+s last updated on 03/May/20 $${let}\:{A}\:{B}\:{C}\:{D}\:{be}\:{four}\:{points}\:{on}\:{a}\:{circle} \\ $$$${let}\:{four}\:{more}\:{cirvces}\:{pass}\:{through}\: \\ $$$${AB}\:,\:{BC}\:,\:{CD}\:,\:{DA}\:{Respectively}, \\ $$$${meeting}\:{infurther}\:{points}\:{P},{Q},{R},{S}. \\ $$$${show}\:{that}\:{P},{Q},{R}\:{and}\:{S}\:{are}\:{consyclic}.…
Question Number 91760 by M±th+et+s last updated on 02/May/20 Commented by Tony Lin last updated on 03/May/20 Commented by M±th+et+s last updated on 03/May/20 $${thanx}\:{for}\:{solutions}…
Question Number 26205 by ajfour last updated on 22/Dec/17 Commented by ajfour last updated on 22/Dec/17 $${Find}\:{the}\:{area}\:{of}\:{the}\:{uncoloured} \\ $$$${region}\:{inside}\:{triangle}\:{in}\:{terms} \\ $$$${of}\:{radii}\:{a}\:,\:{and}\:{b}\:{of}\:{the}\:{lower} \\ $$$${circles}. \\ $$…
Question Number 26162 by lizan 123 last updated on 21/Dec/17 $${Find}\:\:{the}\:{volume}\:\:{of}\:{the}\:{solid}\:{generated}\:{when}\:{the}\:{region}\:\:{enclosed} \\ $$$${by}\:\:{y}=\:\sqrt{{x}} \\ $$$$\:{y}=\mathrm{0}\:\:{and}\:\:{x}=\mathrm{9}\:{about}\:{the}\:\:{line}\:{x}=\mathrm{9} \\ $$ Answered by ajfour last updated on 21/Dec/17 Commented…
Question Number 91568 by jagoll last updated on 01/May/20 Answered by MJS last updated on 01/May/20 $$\mathrm{pipe}\:\mathrm{A}:\:\frac{\mathrm{1}}{\mathrm{20}}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{pipe}\:\mathrm{B}:\:\frac{\mathrm{1}}{\mathrm{30}}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{pipe}\:\mathrm{C}:\:−\frac{\mathrm{1}}{\mathrm{36}}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{8}:\mathrm{00}\:=\mathrm{0} \\ $$$$\mathrm{9}:\mathrm{30}\:=\frac{\mathrm{3}}{\mathrm{2}}…
Question Number 25895 by Mr eaay last updated on 16/Dec/17 $${The}\:{hands}\:{of}\:{a}\:{clock}\:{have}\:{length}\:\mathrm{3}{cm}\: \\ $$$${and}\:\mathrm{5}{cm}.{Calculate}\:{the}\:{distance}\:{between} \\ $$$${the}\:{tips}\:{of}\:{the}\:{hand}\:{at}\:\mathrm{4}\:“\mathrm{0}''\:{clock}. \\ $$ Answered by mrW1 last updated on 16/Dec/17 $${angle}\:{between}\:{the}\:{hands}:…
Question Number 25871 by ajfour last updated on 16/Dec/17 Commented by ajfour last updated on 16/Dec/17 $$\bigtriangleup{ABC}\:{is}\:{equilateral}\:{and}\:{G}\:{its} \\ $$$${centroid}.\:{Parabolas}\:{AGB},\:{BGC}, \\ $$$${and}\:{CGA}\:{each}\:{have}\:{their}\:{vertices} \\ $$$${at}\:{G}.\:{Find}\:{ratio}\:{of}\:{area}\:{in}\:{brown} \\ $$$${to}\:{the}\:{area}\:{of}\:{the}\:\bigtriangleup{ABC}\:.…
Question Number 25718 by ajfour last updated on 13/Dec/17 Commented by mrW1 last updated on 13/Dec/17 $${Case}\:\mathrm{1}:\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${h}_{{i}} =\sqrt{{h}^{\mathrm{2}} +{R}^{\mathrm{2}} } \\ $$$${A}_{{i}} =\frac{\mathrm{2}{R}×{h}_{{i}}…
Question Number 25702 by ajfour last updated on 13/Dec/17 Commented by ajfour last updated on 13/Dec/17 $${solution}\:{to}\:{Q}.\:\mathrm{25700} \\ $$ Answered by ajfour last updated on…