Question Number 65119 by Tawa1 last updated on 25/Jul/19 Commented by Tony Lin last updated on 25/Jul/19 Commented by Tony Lin last updated on 25/Jul/19…
Question Number 64971 by Tawa1 last updated on 23/Jul/19 Commented by Tony Lin last updated on 23/Jul/19 $$\frac{\frac{{x}}{\mathrm{2}}}{{sin}\theta}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{sin}\left(\mathrm{90}°+\theta\right)}=\frac{{x}}{{sin}\left(\mathrm{90}°−\mathrm{2}\theta\right)} \\ $$$$\Rightarrow\frac{\frac{{x}}{\mathrm{2}}}{{sin}\theta}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{cos}\theta}=\frac{{x}}{{cos}\mathrm{2}\theta} \\ $$$${let}\:{cos}\theta={t} \\ $$$$\Rightarrow\frac{\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{t}}=\frac{{x}}{\mathrm{2}{t}^{\mathrm{2}}…
Question Number 64949 by Tony Lin last updated on 23/Jul/19 Commented by Tony Lin last updated on 23/Jul/19 $${find}\:{x} \\ $$ Answered by MJS last…
Question Number 64909 by Tawa1 last updated on 23/Jul/19 Commented by MJS last updated on 23/Jul/19 $$\mathrm{the}\:\mathrm{picture}\:\mathrm{shows}\:\mathrm{2}\:\mathrm{circles},\:\mathrm{so}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{numbers}\:\mathrm{don}'\mathrm{t}\:\mathrm{fit}.\:\mathrm{obviously}\:\mathrm{the}\:\mathrm{right}\:\mathrm{must} \\ $$$$\mathrm{be}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{the}\:\mathrm{above}/\mathrm{below}\:\mathrm{numbers} \\ $$ Commented by…
Question Number 64901 by LPM last updated on 23/Jul/19 Commented by LPM last updated on 23/Jul/19 $$\mathrm{KM}=? \\ $$ Commented by Tony Lin last updated…
Question Number 64892 by mathmax by abdo last updated on 22/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 64887 by ajfour last updated on 22/Jul/19 Answered by mr W last updated on 22/Jul/19 $${let}\:\lambda=\frac{{a}}{{R}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}×\frac{{a}}{\mathrm{2}}×{R}}=\frac{\mathrm{4}{R}^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}{aR}}=\frac{\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}}…
Question Number 130339 by mnjuly1970 last updated on 24/Jan/21 Answered by MJS_new last updated on 25/Jan/21 $${A}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${AC}:\:{y}=−\frac{{a}}{{c}}{x}+{a} \\ $$$$\mathrm{0}\leqslant{d}\in\mathbb{R} \\ $$$${BD}:\:{y}={dx} \\ $$$${D}=\begin{pmatrix}{\frac{{ac}}{{a}+{cd}}}\\{\frac{{acd}}{{a}+{cd}}}\end{pmatrix}…
Question Number 130241 by mr W last updated on 23/Jan/21 Commented by mr W last updated on 23/Jan/21 $${three}\:{circles}\:{with}\:{radii}\:{p},\:{q},\:{r}\:{have} \\ $$$${the}\:{same}\:{center}.\:\left({p}<{q}<{r}\right) \\ $$$${what}\:{is}\:{the}\:{side}\:{length}\:{of}\:{the}\:{smallest} \\ $$$${and}\:{the}\:{largest}\:{square}\:{which}\:{has}…
Question Number 64686 by ajfour last updated on 20/Jul/19 Answered by ajfour last updated on 20/Jul/19 $${Let}\:{B}\:{be}\:{origin}\:\:{and}\:{x}\:{axis}\:{along} \\ $$$${BC}. \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\left({x}−{a}\right)^{\mathrm{2}}…