Question Number 2329 by 123456 last updated on 16/Nov/15 $${a}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \left(\lfloor\frac{{n}}{\mathrm{9}}\rfloor+\mathrm{1}\right) \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}{a}_{{n}} =? \\ $$$${b}_{{n}} =\frac{\left({a}_{{n}} −\mathrm{1}\right)\left(\mathrm{9}−{a}_{{n}} \right)}{{n}+\mathrm{1}} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}{b}_{{n}} =?…
Question Number 67852 by mr W last updated on 01/Sep/19 Commented by Prithwish sen last updated on 01/Sep/19 $$ \\ $$$$\boldsymbol{\mathrm{A}}=\left(\mathrm{0},\boldsymbol{\mathrm{a}}\right),\:\boldsymbol{\mathrm{B}}\:=\:\left(\frac{−\mathrm{4}}{\boldsymbol{\mathrm{a}}},\mathrm{0}\right),\boldsymbol{\mathrm{C}}=\left(\frac{\mathrm{20}}{\boldsymbol{\mathrm{a}}},−\mathrm{9}\boldsymbol{\mathrm{a}}\right),\boldsymbol{\mathrm{D}}=\left(\frac{\mathrm{2}}{\boldsymbol{\mathrm{a}}},\mathrm{0}\right) \\ $$$$\boldsymbol{\mathrm{E}}=\left(\mathrm{0},\frac{−\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}\right)\: \\ $$$$\mathrm{Equation}\:\mathrm{of}\:\mathrm{line}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{A}\:\mathrm{and}\:\mathrm{D}…
Question Number 67849 by mr W last updated on 01/Sep/19 Commented by MJS last updated on 01/Sep/19 $$\mathrm{is}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{90}°? \\ $$ Commented by mr W last…
Question Number 133371 by I want to learn more last updated on 21/Feb/21 Answered by mr W last updated on 22/Feb/21 Commented by mr W…
Question Number 67813 by mr W last updated on 31/Aug/19 Commented by mr W last updated on 31/Aug/19 $${shaded}\:{area}\:{A}=? \\ $$ Commented by Prithwish sen…
Question Number 67807 by TawaTawa last updated on 31/Aug/19 Commented by MJS last updated on 01/Sep/19 $$\mathrm{perimeter}\:\mathrm{or}\:\mathrm{area}? \\ $$ Commented by TawaTawa last updated on…
Question Number 67774 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${DE}={AB}=\mathrm{2}{r} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{semicircle}\:=\frac{\pi}{\mathrm{2}}{r}^{\mathrm{2}} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:=\frac{\mathrm{1}}{\mathrm{2}}\mid{AB}\mid{h}={rh} \\ $$$$\:\:\:\:\:\mathrm{with}\:{h}={r}\mathrm{tan}\:{x} \\ $$$$\:\:\:\:\:={r}^{\mathrm{2}}…
Question Number 67772 by TawaTawa last updated on 31/Aug/19 Commented by Prithwish sen last updated on 31/Aug/19 $$\boldsymbol{\mathrm{r}}=\boldsymbol{\mathrm{a}}\left(\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$ Commented by Prithwish sen last…
Question Number 67775 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{side}\:{s}=\sqrt{\mathrm{196}}=\mathrm{14} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{sector}'\mathrm{s}\:\mathrm{area}\:=\frac{\pi}{\mathrm{8}}{s}^{\mathrm{2}} =\frac{\mathrm{49}\pi}{\mathrm{2}}\:\mathrm{minus} \\ $$$$\mathrm{the}\:\mathrm{white}\:\mathrm{segment}\:\mathrm{which}\:\mathrm{intersects}\:\mathrm{the} \\ $$$$\mathrm{diagonal}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square} \\…
Question Number 67770 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${AD}+{DE}={AE}=\mathrm{6} \\ $$$$\Rightarrow\:{AD}={q};\:{DE}=\mathrm{6}−{q}\:\:\left[\Rightarrow\:{q}<\mathrm{6}\right] \\ $$$${AB}={DE}−{AD}=\mathrm{6}−\mathrm{2}{q}\:\:\left[\Rightarrow\:{q}<\mathrm{3}\right] \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{{q}}\\{\mathrm{0}}\end{pmatrix}\:\:{E}=\begin{pmatrix}{\mathrm{6}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{line}\:{AB}:\:{y}={x}\mathrm{tan}\:\mathrm{60}°\:\Leftrightarrow\:{y}=\sqrt{\mathrm{3}}{x}\:\:\left[\mathrm{1}\right]…