Question Number 157292 by mr W last updated on 21/Oct/21 Commented by mr W last updated on 21/Oct/21 Commented by mr W last updated on…
Question Number 157225 by mr W last updated on 21/Oct/21 Commented by mr W last updated on 21/Oct/21 $${Solution}\:{to}\:{Q}\mathrm{157114} \\ $$$$ \\ $$$${AD}=\sqrt{\mathrm{15}^{\mathrm{2}} −\left(\mathrm{3}{x}\right)^{\mathrm{2}} }…
Question Number 157114 by amin96 last updated on 20/Oct/21 Commented by mathlove last updated on 20/Oct/21 $${wdzr}\:{from}\:{yoj} \\ $$ Commented by mathlove last updated on…
Question Number 157054 by andradamaryjane last updated on 19/Oct/21 $${suppose}\:{you}\:{drop}\:{a}\:{tennis}\:{ball}\:{from}\:{a}\:{hieght}\:{of}\:\mathrm{15}\:{feet}.{after}\:{the}\:{ballhits}\:{the}\:{floor}\:{it}\:{rebounds}\:\:{to}\mathrm{85\%}\:{of}\:{its}\:{previous}\:{height}.{how}\:{high}\:{will}\:{the}\:{ball}\:{rebound}\:{after}\:{its}\:{ghird}\:{bounce}\:{round}\:{tl}\:{the}\:{nearest}\:{tenth} \\ $$$$ \\ $$ Commented by mr W last updated on 19/Oct/21 $${suppose}\:{you}\:{don}'{t}\:{write}\:{all}\:{text}\:{in}\:{a} \\ $$$${single}\:{line},\:{maybe}\:{more}\:{people}\:{may}…
Question Number 25963 by tawa tawa last updated on 16/Dec/17 $$\mathrm{A}\:\mathrm{bus}\:\mathrm{is}\:\mathrm{traveling}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}\:\mathrm{at}\:\mathrm{100}\:\mathrm{km}/\mathrm{hr}\:\mathrm{and}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{conductor} \\ $$$$\mathrm{walks}\:\mathrm{at}\:\mathrm{6}\:\mathrm{km}/\mathrm{hr}\:\mathrm{on}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\mathrm{as}\:\mathrm{the}\:\mathrm{bus}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{conductor}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{road}\:\mathrm{and}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bus}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 25930 by Tinkutara last updated on 16/Dec/17 $$\mathrm{A}\:\mathrm{line}\:\mathrm{passes}\:\mathrm{through}\:{A}\left(−\mathrm{3},\:\mathrm{0}\right)\:\mathrm{and} \\ $$$${B}\left(\mathrm{0},\:−\mathrm{4}\right).\:\mathrm{A}\:\mathrm{variable}\:\mathrm{line}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:{AB}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{cut}\:{x}\:\mathrm{and}\:{y}-\mathrm{axes}\:\mathrm{at} \\ $$$${M}\:\mathrm{and}\:{N}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lines}\:{AN}\:\mathrm{and}\:{BM}. \\ $$ Answered by ajfour last updated…
Question Number 156940 by amin96 last updated on 17/Oct/21 Answered by mr W last updated on 17/Oct/21 Commented by mr W last updated on 17/Oct/21…
Question Number 91275 by I want to learn more last updated on 29/Apr/20 Commented by mr W last updated on 29/Apr/20 $$\left({A}\right) \\ $$ Commented…
Question Number 25609 by behi.8.3.4.17@gmail.com last updated on 12/Dec/17 Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17 $$\boldsymbol{\mathrm{BE}}=\boldsymbol{\mathrm{EC}},\boldsymbol{\mathrm{AB}}=\mathrm{12},\boldsymbol{\mathrm{AC}}=\mathrm{10} \\ $$$$\boldsymbol{\mathrm{parallel}}\:\boldsymbol{\mathrm{lines}}\:\boldsymbol{\mathrm{to}}:\boldsymbol{\mathrm{AE}},\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{equal}}\:\boldsymbol{\mathrm{distance}} \\ $$$$\boldsymbol{\mathrm{from}}:\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{B}},\boldsymbol{\mathrm{toward}}\:\boldsymbol{\mathrm{C}},\boldsymbol{\mathrm{divide}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{area}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{A}}\overset{\bigtriangleup} {\boldsymbol{\mathrm{B}C}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{ratio}}:\:\mathrm{1}:\mathrm{4}:\mathrm{2}:\mathrm{3}\:. \\…
Question Number 25605 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17 Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17 $$\mathrm{from}\:\mathrm{midpoints}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{draw}\:\mathrm{perpendicular} \\ $$$$\mathrm{lines}\:\mathrm{to}\:\mathrm{opposite}\:\mathrm{sides}.\mathrm{find}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{inner}\:\mathrm{hexagon}\:\mathrm{to}\:\mathrm{area}\:\mathrm{of}:\mathrm{A}\overset{\bigtriangleup} {\mathrm{B}C}. \\ $$ Commented…