Question Number 65022 by ajfour last updated on 24/Jul/19 Commented by MJS last updated on 24/Jul/19 $$\mathrm{I}\:\mathrm{used}\:\mathrm{a}\:\mathrm{calculator}.\:\mathrm{I}'\mathrm{ve}\:\mathrm{got}\:\mathrm{an}\:\mathrm{old}\:\mathrm{TI}−\mathrm{89} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{any}\:\mathrm{calculator}\:\mathrm{which}\:\mathrm{is}\:\mathrm{able}\:\mathrm{to} \\ $$$$\mathrm{approximately}\:\mathrm{solve}\:\mathrm{equations}\:\mathrm{and}\:\mathrm{integrals} \\ $$$$\mathrm{select}\:\mathrm{a}\:\mathrm{value}\:\mathrm{for}\:{r} \\ $$$$\mathrm{calculate}\:{P},\:{Q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{integrals}…
Question Number 64968 by ajfour last updated on 23/Jul/19 Commented by ajfour last updated on 25/Jul/19 $${Instead}\:{let}\:\:{y}={Ax}^{\mathrm{2}} \:,\:{find}\:{side} \\ $$$${length}\:{of}\:{such}\:{a}\:{rhombus}\:{that} \\ $$$${fits}\:{in},\:{the}\:{way}\:{shown}.\:{Also} \\ $$$${find}\:{A}.\:\:\:\:\:\:\: \\…
Question Number 130358 by ajfour last updated on 24/Jan/21 Commented by ajfour last updated on 24/Jan/21 $${Two}\:{identical}\:{ellipses}. \\ $$$${Find}\:{maximum}\:{value}\:{of}\:{b}/{a}. \\ $$ Commented by MJS_new last…
Question Number 130263 by pete last updated on 23/Jan/21 Answered by Olaf last updated on 23/Jan/21 $$\mathrm{B}\begin{pmatrix}{{x}_{\mathrm{B}} \:=\:\mathrm{12}−\mathrm{2}{y}_{\mathrm{B}} \:=\:\mathrm{12}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{E}\begin{pmatrix}{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{{y}_{\mathrm{E}} }{{x}_{\mathrm{B}} }\:\Rightarrow\:{y}_{\mathrm{E}} \:=\:\mathrm{6}}\end{pmatrix} \\…
Question Number 64709 by ajfour last updated on 20/Jul/19 Commented by ajfour last updated on 20/Jul/19 $${If}\:{regions}\:{A}\:{and}\:{B}\:{have}\:{the}\:{same} \\ $$$${area},\:{and}\:{both}\:{parabolas}\:{have} \\ $$$${the}\:{same}\:{shape}\:{but}\:{their}\:{axis}\:\bot\:{to} \\ $$$${each}\:{other},\:{then}\:{determine} \\ $$$${the}\:{equation}\:{of}\:{the}\:{red}\:{parabola}.…
Question Number 64436 by ajfour last updated on 17/Jul/19 Commented by ajfour last updated on 17/Jul/19 $${Find}\:{radius}\:{of}\:{arc}\:{length}\:{s}\:,\:{if} \\ $$$${area}\:{between}\:{parabola}\:{and}\:{arc} \\ $$$${is}\:{a}\:{maximum}. \\ $$ Answered by…
Question Number 64311 by aseer imad last updated on 16/Jul/19 Commented by Tony Lin last updated on 17/Jul/19 $${letA}'={A}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{0},\mathrm{0}\right),{C}'={C}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$${D}'=\begin{bmatrix}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{cos}\frac{\pi}{\mathrm{4}}\:\:−{sin}\frac{\pi}{\mathrm{4}}}\\{{sin}\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:{cos}\frac{\pi}{\mathrm{4}}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:=\begin{bmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{\:\:\:\:\frac{\mathrm{9}}{\mathrm{2}}}\end{bmatrix}…
Question Number 64309 by aseer imad last updated on 16/Jul/19 Commented by Tony Lin last updated on 17/Jul/19 $${let}\:{A}'={A}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{0},\mathrm{0}\right),{C}'={C}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$${D}'\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{5}{i}\right)\left({cos}\frac{\pi}{\mathrm{4}}+{isin}\frac{\pi}{\mathrm{4}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}{i} \\ $$$$\Rightarrow{D}=\left(\mathrm{1},\mathrm{3}\right)+\left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{9}}{\mathrm{2}}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right) \\ $$$${B}'\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{5}{i}\right)\left({cos}\frac{\pi}{\mathrm{4}}−{isin}\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i}…
Question Number 129436 by mr W last updated on 07/Feb/21 Commented by mr W last updated on 07/Feb/21 $${see}\:{Q}\mathrm{129244} \\ $$ Commented by ajfour last…
Question Number 129404 by bemath last updated on 15/Jan/21 $$\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:\theta−{i}\:\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}}.\mathrm{cot}\:\left(\frac{\theta}{\mathrm{2}}\right) \\ $$$$\mathrm{prove}. \\ $$ Answered by MJS_new last updated on 15/Jan/21 $$\theta=\mathrm{2}\alpha \\ $$$$\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\alpha\:−\mathrm{i}\:\mathrm{sin}\:\mathrm{2}\alpha\right)^{−\mathrm{1}} =…