Question Number 133316 by bramlexs22 last updated on 21/Feb/21 $$\mathrm{How}\:\mathrm{many}\:\mathrm{6}−\mathrm{letter}\:\mathrm{words}\:\mathrm{in}\: \\ $$$$\mathrm{which}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{letter}\:\mathrm{appears} \\ $$$$\mathrm{more}\:\mathrm{than}\:\mathrm{once}\:,\mathrm{can}\:\mathrm{be}\:\mathrm{made}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{letters}\:\mathrm{in}\:\mathrm{the}\:\mathrm{word}\:\mathrm{FLIGHT} \\ $$$$ \\ $$ Commented by mr W last…
Question Number 133318 by bramlexs22 last updated on 21/Feb/21 $$\left(\mathrm{a}\right)\:\mathrm{Are}\:\mathrm{there}\:\mathrm{any}\:\mathrm{graphs}\:\mathrm{with}\:\mathrm{5} \\ $$$$\mathrm{vertices}\:\mathrm{which}\:\mathrm{have}\:\mathrm{vertices}\:\mathrm{of}\: \\ $$$$\mathrm{degrees}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\:\mathrm{and}\:\mathrm{5}?\: \\ $$ Answered by EDWIN88 last updated on 21/Feb/21 $$\mathrm{No};\:\mathrm{if}\:\mathrm{a}\:\mathrm{graph}\:\mathrm{has}\:\mathrm{n}\:\mathrm{vertices}\:\mathrm{it}\:\mathrm{can}\:\mathrm{have}\:\mathrm{no}\: \\…
Question Number 2241 by Yozzi last updated on 10/Nov/15 $${Evaluate} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{9}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\frac{{sin}\theta{cos}\theta}{{cos}^{\mathrm{3}} \theta+{sin}^{\mathrm{3}} \theta}\right)^{\mathrm{2}} {d}\theta. \\ $$ Answered by Filup last updated on…
Question Number 133314 by bramlexs22 last updated on 21/Feb/21 $$\mathrm{How}\:\mathrm{many}\:\mathrm{rearrangements}\:\mathrm{are}\: \\ $$$$\mathrm{there}\:\mathrm{of}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{in}\:\mathrm{the}\:\mathrm{world} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{ENGINEERING} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{MATHEMATICAL}\: \\ $$ Answered by EDWIN88 last updated on 21/Feb/21…
Question Number 2240 by Rasheed Soomro last updated on 10/Nov/15 $$\mathcal{GENERALIZE}: \\ $$$$\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)={a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$$$\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}}…
Question Number 67774 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${DE}={AB}=\mathrm{2}{r} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{semicircle}\:=\frac{\pi}{\mathrm{2}}{r}^{\mathrm{2}} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:=\frac{\mathrm{1}}{\mathrm{2}}\mid{AB}\mid{h}={rh} \\ $$$$\:\:\:\:\:\mathrm{with}\:{h}={r}\mathrm{tan}\:{x} \\ $$$$\:\:\:\:\:={r}^{\mathrm{2}}…
Question Number 67775 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{side}\:{s}=\sqrt{\mathrm{196}}=\mathrm{14} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{sector}'\mathrm{s}\:\mathrm{area}\:=\frac{\pi}{\mathrm{8}}{s}^{\mathrm{2}} =\frac{\mathrm{49}\pi}{\mathrm{2}}\:\mathrm{minus} \\ $$$$\mathrm{the}\:\mathrm{white}\:\mathrm{segment}\:\mathrm{which}\:\mathrm{intersects}\:\mathrm{the} \\ $$$$\mathrm{diagonal}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square} \\…
Question Number 67772 by TawaTawa last updated on 31/Aug/19 Commented by Prithwish sen last updated on 31/Aug/19 $$\boldsymbol{\mathrm{r}}=\boldsymbol{\mathrm{a}}\left(\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$ Commented by Prithwish sen last…
Question Number 133305 by liberty last updated on 21/Feb/21 $$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\mathrm{5}+\mathrm{3sin}\:\mathrm{2x}}\:=? \\ $$ Answered by physicstutes last updated on 21/Feb/21 $$\mathrm{set}\:{t}\:=\:\mathrm{tan}\:{x} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:=\:\frac{\mathrm{2}\:\mathrm{tan}\:{x}}{\mathrm{1}+\:\mathrm{tan}^{\mathrm{2}} {x}}\:\:=\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 67770 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${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]…