Question Number 64557 by LPM last updated on 19/Jul/19 Commented by LPM last updated on 19/Jul/19 $$\mathrm{area}\:\mathrm{of}\:\bigtriangleup\:\mathrm{DBC}\: \\ $$$$ \\ $$ Commented by Tony Lin…
Question Number 64545 by ajfour last updated on 19/Jul/19 Commented by ajfour last updated on 19/Jul/19 Commented by aliesam last updated on 19/Jul/19 Answered by…
Question Number 64544 by LPM last updated on 19/Jul/19 Commented by LPM last updated on 19/Jul/19 $${area}\:{of}\:\:\Box{ABCD} \\ $$ Answered by MJS last updated on…
Question Number 130067 by Study last updated on 22/Jan/21 Answered by MJS_new last updated on 22/Jan/21 $$\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{7}{x}+\mathrm{3}{y}−\mathrm{21}=\mathrm{0} \\ $$$$\mathrm{intersects}\:\mathrm{the}\:{x}−\mathrm{axis}\:\mathrm{at}\:{a}=\mathrm{3}\:\mathrm{and}\:\mathrm{the}\:{y}−\mathrm{axis} \\ $$$$\mathrm{at}\:{b}=\mathrm{7}\:\mathrm{and}\:{a}\neq{b}\neq\mathrm{0}\:\mathrm{and}\:\mathrm{both}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{opposite}\:\mathrm{to}\:\mathrm{zero}…
Question Number 130065 by Study last updated on 22/Jan/21 Commented by MJS_new last updated on 22/Jan/21 $$\mathrm{how}\:\mathrm{could}\:\mathrm{this}\:\mathrm{be}\:\mathrm{true}?\:\mathrm{it}'\mathrm{s}\:\mathrm{nonsense} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{line}\:\mathrm{through}\:\begin{pmatrix}{{a}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{and}\:\begin{pmatrix}{\mathrm{0}}\\{{b}}\end{pmatrix}\:\mathrm{for}\:\mathrm{sny} \\ $$$$\mathrm{given}\:\mathrm{pair}\:\mathrm{of}\:{a},\:{b}\:\in\mathbb{R}\:\mathrm{with}\:\mathrm{not}\:\left({a}=\mathrm{0}\wedge{b}=\mathrm{0}\right) \\ $$ Terms of…
Question Number 64516 by Tawa1 last updated on 18/Jul/19 Commented by Tawa1 last updated on 18/Jul/19 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{area}\:\mathrm{of}\:\mathrm{each}\:\mathrm{colour}\:? \\ $$$$\mathrm{please}\:\mathrm{help}. \\ $$ Commented by Tawa1 last…
Question Number 64469 by Tawa1 last updated on 18/Jul/19 Answered by MJS last updated on 18/Jul/19 $$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{incircle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangular} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${r}=\frac{{a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\frac{{a}+{b}−\sqrt{{a}^{\mathrm{2}}…
Question Number 64459 by Tawa1 last updated on 18/Jul/19 Commented by Tony Lin last updated on 18/Jul/19 $${AR}={AP}=\mathrm{2},{RC}={QC}=\mathrm{5} \\ $$$${letPB}={BQ}={x} \\ $$$${cos}\mathrm{60}°=\frac{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} +\left(\mathrm{5}+{x}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}+{x}\right)\left(\mathrm{5}+{x}\right)}…
Question Number 64425 by Tawa1 last updated on 17/Jul/19 Commented by Prithwish sen last updated on 17/Jul/19 $$\mathrm{Area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{portion}\:=\: \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{square}\:−\:\mathrm{Area}\:\mathrm{of}\:\mathrm{quarter}\:\mathrm{circle} \\ $$$$=\:\mathrm{a}^{\mathrm{2}} \:−\:\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\:\left[\:\mathrm{if}\:\boldsymbol{\mathrm{a}}\:\mathrm{be}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\right. \\…
Question Number 129910 by Algoritm last updated on 20/Jan/21 Commented by mr W last updated on 20/Jan/21 $${no}\:{unique}\:{solution}!\: \\ $$$${please}\:{make}\:{your}\:{graph}\:{clear}! \\ $$ Answered by mr…