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Question Number 190216    Answers: 2   Comments: 3

Question Number 190167    Answers: 0   Comments: 0

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Question Number 190052    Answers: 1   Comments: 0

Evaluate ∫∫_A (x+y)^2 dxdy over the area bounded by the ellipse (x^2 /a^2 ) + (y^2 /b^2 ) = 1 Anybody?

$$\mathrm{Evaluate}\:\int\int_{\mathrm{A}} \left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \mathrm{dxdy}\:\mathrm{over}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{ellipse}\: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Anybody}? \\ $$

Question Number 189865    Answers: 0   Comments: 2

i know b_n =0, but a_0 =(4/3)F_0 (according to solution) my answer is a_0 =(8/3)F_0 where did i did wrong how to find answer (Fourier transformation) like picture below (in the comment? f(t)= { ((((3F_0 )/t_0 )t,0≤t≤(t_0 /3))),((F_0 , (t_0 /3)≤t≤((2t_0 )/3))),((((-3F_0 )/t_0 )t+3F_0 ,((2t_0 )/3)≤t≤t_0 )) :} a_n =(2/t_0 )∫_0 ^t_0 f(t) cos(nωt)dt a_n =(2/t_0 )∫_0 ^t_0 f(t) cos(nωt)dt a_0 =(2/t_0 )∫_0 ^t_0 f(t) dt ∫_0 ^((4t_0 )/3) f(t) dt =∫_0 ^(t_0 /3) ((3F_0 )/t_0 )tdt+∫_(t_0 /3) ^((2t_0 )/3) F_0 dt+∫_((2t_0 )/3) ^t_0 ((-3F_0 )/t_0 )t+3F_0 dt =[((3F_0 )/t_0 )t]_0 ^(t_0 /3) +[F_0 t]_(t_0 /3) ^((2t_0 )/3) +[((-3F_0 )/t_0 )t+3F_0 t]_((2t_0 )/3) ^t_0 =F_0 +(1/3)F_0 t_0 −F_0 +F_0 t_0 =(4/3) F_0 t_0 so a_0 =(2/t_0 )×(4/3)F_0 t_0 =(8/3)F_0

$$ \\ $$$$\:{i}\:{know}\:{b}_{{n}} =\mathrm{0},\:{but}\:{a}_{\mathrm{0}} =\frac{\mathrm{4}}{\mathrm{3}}{F}_{\mathrm{0}} \left({according}\:\right. \\ $$$$\left.\:{to}\:{solution}\right)\:{my}\:{answer}\:{is}\:{a}_{\mathrm{0}} =\frac{\mathrm{8}}{\mathrm{3}}{F}_{\mathrm{0}} \\ $$$$\:{where}\:{did}\:{i}\:{did}\:{wrong}\:{how}\:{to} \\ $$$${find}\:{answer}\:\left({Fourier}\:{transformation}\right) \\ $$$$\:{like}\:{picture}\:{below}\:\left({in}\:{the}\:{comment}?\right. \\ $$$$\:{f}\left({t}\right)=\begin{cases}{\frac{\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t},\mathrm{0}\leqslant{t}\leqslant\frac{{t}_{\mathrm{0}} }{\mathrm{3}}}\\{{F}_{\mathrm{0}} ,\:\frac{{t}_{\mathrm{0}} }{\mathrm{3}}\leqslant{t}\leqslant\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}}\\{\frac{-\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t}+\mathrm{3}{F}_{\mathrm{0}} ,\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}\leqslant{t}\leqslant{t}_{\mathrm{0}} }\end{cases} \\ $$$$\:{a}_{{n}} =\frac{\mathrm{2}}{{t}_{\mathrm{0}} }\int_{\mathrm{0}} ^{{t}_{\mathrm{0}} } {f}\left({t}\right)\:{cos}\left({n}\omega{t}\right){dt}\: \\ $$$$\:{a}_{{n}} =\frac{\mathrm{2}}{{t}_{\mathrm{0}} }\int_{\mathrm{0}} ^{{t}_{\mathrm{0}} } {f}\left({t}\right)\:{cos}\left({n}\omega{t}\right){dt}\: \\ $$$$\:{a}_{\mathrm{0}} =\frac{\mathrm{2}}{{t}_{\mathrm{0}} }\int_{\mathrm{0}} ^{{t}_{\mathrm{0}} } {f}\left({t}\right)\:{dt}\: \\ $$$$\:\int_{\mathrm{0}} ^{\frac{\mathrm{4}{t}_{\mathrm{0}} }{\mathrm{3}}} {f}\left({t}\right)\:{dt}\:=\int_{\mathrm{0}} ^{\frac{{t}_{\mathrm{0}} }{\mathrm{3}}} \frac{\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{tdt}+\int_{\frac{{t}_{\mathrm{0}} }{\mathrm{3}}} ^{\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}} {F}_{\mathrm{0}} {dt}+\int_{\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}} ^{{t}_{\mathrm{0}} } \frac{-\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t}+\mathrm{3}{F}_{\mathrm{0}} {dt} \\ $$$$\:\:=\left[\frac{\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t}\right]_{\mathrm{0}} ^{\frac{{t}_{\mathrm{0}} }{\mathrm{3}}} +\left[{F}_{\mathrm{0}} {t}\right]_{\frac{{t}_{\mathrm{0}} }{\mathrm{3}}} ^{\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}} +\left[\frac{-\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t}+\mathrm{3}{F}_{\mathrm{0}} {t}\right]_{\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}} ^{{t}_{\mathrm{0}} } \\ $$$$\:={F}_{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{3}}{F}_{\mathrm{0}} {t}_{\mathrm{0}} −{F}_{\mathrm{0}} +{F}_{\mathrm{0}} {t}_{\mathrm{0}} \\ $$$$\:=\frac{\mathrm{4}}{\mathrm{3}}\:{F}_{\mathrm{0}} {t}_{\mathrm{0}} \\ $$$$\:{so}\:{a}_{\mathrm{0}} =\frac{\mathrm{2}}{{t}_{\mathrm{0}} }×\frac{\mathrm{4}}{\mathrm{3}}{F}_{\mathrm{0}} {t}_{\mathrm{0}} =\frac{\mathrm{8}}{\mathrm{3}}{F}_{\mathrm{0}} \\ $$$$ \\ $$$$ \\ $$

Question Number 189844    Answers: 0   Comments: 0

lim_(x→∞) [xsin((π/x))] Hi

$$\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\left[\mathrm{xsin}\left(\frac{\pi}{\mathrm{x}}\right)\right] \\ $$$$ \\ $$$$\mathrm{Hi} \\ $$

Question Number 189755    Answers: 0   Comments: 0

Question Number 189754    Answers: 0   Comments: 0

Question Number 189753    Answers: 0   Comments: 0

Question Number 189716    Answers: 3   Comments: 0

Question Number 189710    Answers: 3   Comments: 0

Question Number 189375    Answers: 3   Comments: 0

show that : (1/(cscx + cot x)) = cscx − cot x

$$ \\ $$$$\:\:\:\:{show}\:{that}\:: \\ $$$$\:\:\:\:\frac{\mathrm{1}}{{cscx}\:+\:{cot}\:{x}}\:=\:{cscx}\:−\:{cot}\:{x} \\ $$$$ \\ $$$$ \\ $$

Question Number 189292    Answers: 0   Comments: 2

Question Number 189127    Answers: 3   Comments: 0

Question Number 189112    Answers: 0   Comments: 3

Question Number 188954    Answers: 0   Comments: 1

Solve : sin(x) + x^2 + tan(x) = 3x Hey!

$$\mathrm{Solve}\:: \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{tan}\left(\mathrm{x}\right)\:=\:\mathrm{3x} \\ $$$$ \\ $$$$\mathrm{Hey}! \\ $$

Question Number 188937    Answers: 0   Comments: 2

A body of mass 20kg lies on a horizontal floor and is acted on by constant forces of magnitude 100N and mN in the directions 180° and 060° respectively, where m is a constant. The body begins to move in the direction 090°. If there's also a resistance of 1N/5 per kilogram to the motion of the body; (a) calculate the value of m and the magnitude of the resultant force acting on the body (b) find the acceleration of the body.

A body of mass 20kg lies on a horizontal floor and is acted on by constant forces of magnitude 100N and mN in the directions 180° and 060° respectively, where m is a constant. The body begins to move in the direction 090°. If there's also a resistance of 1N/5 per kilogram to the motion of the body; (a) calculate the value of m and the magnitude of the resultant force acting on the body (b) find the acceleration of the body.

Question Number 188927    Answers: 1   Comments: 0

Question Number 188926    Answers: 1   Comments: 0

Question Number 188730    Answers: 2   Comments: 2

let p(x) = (5/3)−6x−9x^2 and Q(y) = −4y^2 −4y+((13)/2) if there exist unique pair of real number (x,y) such that p(x)×Q(y) = 20 then find the value 6x+10y = ?

$$\:\:\mathrm{let}\:{p}\left({x}\right)\:=\:\frac{\mathrm{5}}{\mathrm{3}}−\mathrm{6}{x}−\mathrm{9}{x}^{\mathrm{2}} \:\mathrm{and}\:{Q}\left({y}\right)\:=\:−\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}+\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{unique}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{real}\:\mathrm{number} \\ $$$$\:\left({x},{y}\right)\:\mathrm{such}\:\mathrm{that}\:{p}\left({x}\right)×{Q}\left({y}\right)\:=\:\mathrm{20}\:\mathrm{then} \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{6}{x}+\mathrm{10}{y}\:=\:? \\ $$

Question Number 188657    Answers: 3   Comments: 0

Question Number 188351    Answers: 2   Comments: 0

(√x)+(√(3x−2))=x^2 +1

$$\sqrt{{x}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}={x}^{\mathrm{2}} +\mathrm{1} \\ $$

Question Number 188218    Answers: 1   Comments: 0

Question Number 188203    Answers: 1   Comments: 0

Question Number 188107    Answers: 2   Comments: 0

Question Number 187835    Answers: 5   Comments: 1

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