Question Number 7608 by Tawakalitu. last updated on 06/Sep/16 $${If}\:\:{x}\:=\:{cos}\Theta\:−\:{sin}\Theta \\ $$$${and}\:\:{y}\:=\:{cos}\mathrm{2}\Theta \\ $$$${Show}\:{that},\:\: \\ $$$${y}\:=\:{x}\sqrt{\mathrm{2}\:−\:{x}^{\mathrm{2}} } \\ $$ Commented by sou1618 last updated on…
Question Number 138673 by KwesiDerek last updated on 16/Apr/21 $$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}=\mathrm{72} \\ $$$$\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}=\mathrm{30} \\ $$$$\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}=\mathrm{4} \\ $$$$\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}=? \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 73137 by behi83417@gmail.com last updated on 06/Nov/19 Commented by behi83417@gmail.com last updated on 06/Nov/19 $$\mathrm{A}\overset{\blacktriangle} {\mathrm{B}C},\:\mathrm{is}\:\mathrm{equilateral}\:\mathrm{with}:\:\mathrm{AB}=\mathrm{a} \\ $$$$\mathrm{BDFG},\mathrm{is}\:\mathrm{rectangle}. \\ $$$$\mathrm{find}:\:\mathrm{S}_{\mathrm{B}\overset{\blacktriangle} {\mathrm{H}F}} ,\mathrm{in}\:\mathrm{terms}\:\mathrm{of}:\:\:\mathrm{a}\:. \\…
Question Number 7600 by Rohit last updated on 05/Sep/16 $${solve}\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\mid<\mathrm{3}\:\:{give}\:{solution} \\ $$ Answered by Yozzia last updated on 05/Sep/16 $$\mid{u}\left({x}\right)\mid<{a}\Rightarrow−{a}<{u}\left({x}\right)<{a}. \\ $$$$\therefore\mid\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 7598 by Rohit last updated on 05/Sep/16 $${solve}\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\mid<\mathrm{3} \\ $$ Answered by Rasheed Soomro last updated on 06/Sep/16 $$\mid\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\mid<\mathrm{3}…
Question Number 73131 by behi83417@gmail.com last updated on 06/Nov/19 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\:\:\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\:. \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}^{\mathrm{2}} \\ $$ Commented by mr W last updated on 06/Nov/19 $${x}\geqslant\mathrm{0}…
Question Number 138661 by mr W last updated on 16/Apr/21 Commented by mr W last updated on 16/Apr/21 $${proof}\:{for}\:{Q}\mathrm{138519} \\ $$ Commented by mr W…
Question Number 7590 by FilupSmith last updated on 05/Sep/16 $${S}\:=\:\underset{{n}=\mathrm{2}} {\overset{{k}} {\sum}}\:\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)} \\ $$$${S}=? \\ $$ Commented by Yozzia last updated on 05/Sep/16 $$\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)}\equiv\frac{{a}}{{n}}+\frac{{b}}{{n}−\mathrm{1}} \\…
Question Number 7587 by ten last updated on 05/Sep/16 $${cos}\mathrm{4}{xsin}\mathrm{4}{x} \\ $$ Answered by Rasheed Soomro last updated on 05/Sep/16 $${cos}\mathrm{4}{xsin}\mathrm{4}{x} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\left(\mathrm{1}/\mathrm{2}\right)\left[\mathrm{sin}\left(\alpha+\beta\right)−\mathrm{sin}\left(\alpha−\beta\right)\right]\:\: \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{4}{x}=\left(\mathrm{1}/\mathrm{2}\right)\left[\mathrm{sin}\left(\mathrm{4}{x}+\mathrm{4}{x}\right)−\mathrm{sin}\left(\mathrm{4}{x}−\mathrm{4}{x}\right)\right]\:\:…
Question Number 138656 by ajfour last updated on 16/Apr/21 $${I}=\int\frac{{dx}}{\left({px}+{q}\right)\sqrt{{ax}^{\mathrm{2}} +{bx}+{c}}} \\ $$ Answered by Ar Brandon last updated on 16/Apr/21 $$\mathcal{I}=\int\frac{\mathrm{dx}}{\left(\mathrm{px}+\mathrm{q}\right)\sqrt{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}} \\ $$$$\mathrm{u}=\frac{\mathrm{1}}{\mathrm{px}+\mathrm{q}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{up}}−\frac{\mathrm{q}}{\mathrm{p}}\Rightarrow\mathrm{du}=−\frac{\mathrm{p}}{\left(\mathrm{px}+\mathrm{q}\right)^{\mathrm{2}}…