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AllQuestion and Answers: Page 302

Question Number 189901    Answers: 2   Comments: 0

Question Number 189899    Answers: 1   Comments: 0

Question Number 189898    Answers: 1   Comments: 0

Question Number 189897    Answers: 1   Comments: 0

Question Number 189896    Answers: 1   Comments: 0

Question Number 189895    Answers: 1   Comments: 0

Question Number 189894    Answers: 1   Comments: 0

Question Number 189891    Answers: 2   Comments: 1

Question Number 189890    Answers: 0   Comments: 0

Question Number 189886    Answers: 0   Comments: 0

e^( x) > 1+ x (∀ x >0 ) set: x=(√(π/e)) −1 e^( ((√π)/( (√e))) −1) > ((√π)/( (√e))) ⇒ e^( ((√π)/( (√e)))) > (√π) ( e^( ((√π)/( (√e)))) )^( (√e)) > (√π)^( (√e)) ⇒ e^( (√π)) > (√(π ))^( (√e))

$$ \\ $$$$\:\:{e}^{\:{x}} \:>\:\mathrm{1}+\:{x}\:\:\:\:\left(\forall\:{x}\:>\mathrm{0}\:\right) \\ $$$$\:\:\:{set}:\:{x}=\sqrt{\frac{\pi}{{e}}}\:−\mathrm{1} \\ $$$$\:\:\:\:{e}^{\:\frac{\sqrt{\pi}}{\:\sqrt{{e}}}\:−\mathrm{1}} >\:\frac{\sqrt{\pi}}{\:\sqrt{{e}}}\:\Rightarrow\:{e}^{\:\frac{\sqrt{\pi}}{\:\sqrt{{e}}}} \:>\:\sqrt{\pi}\: \\ $$$$\:\:\:\:\:\:\left(\:{e}^{\:\frac{\sqrt{\pi}}{\:\sqrt{{e}}}} \right)^{\:\sqrt{{e}}} >\:\sqrt{\pi}\:^{\:\sqrt{{e}}} \:\Rightarrow\:{e}^{\:\sqrt{\pi}} \:>\:\sqrt{\pi\:}\:^{\:\sqrt{{e}}} \\ $$$$ \\ $$

Question Number 189874    Answers: 2   Comments: 0

∫_0 ^( 2) ((sin(x))/(sin(x)+cos(x)))dx= ?

$$\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\:\:\frac{{sin}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}=\:? \\ $$

Question Number 189873    Answers: 1   Comments: 0

∫_0 ^( 2) ((x−1)/(x^2 +x+1))dx= ?

$$\int_{\mathrm{0}} ^{\:\mathrm{2}} \frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=\:? \\ $$

Question Number 189872    Answers: 1   Comments: 0

A farmer and his son used 20 days to weed a plot of land . If the farmer uses 36 days to weed the same plot of land , how many days will the son use to weed the same plot of land

$${A}\:{farmer}\:{and}\:{his}\:{son}\:{used}\:\mathrm{20}\:{days}\: \\ $$$${to}\:{weed}\:{a}\:{plot}\:{of}\:{land}\:.\:{If}\:{the}\:{farmer} \\ $$$${uses}\:\mathrm{36}\:{days}\:{to}\:{weed}\:{the}\:{same}\:{plot}\:{of} \\ $$$${land}\:,\:{how}\:{many}\:{days}\:{will}\:{the}\:{son}\: \\ $$$${use}\:{to}\:{weed}\:{the}\:{same}\:{plot}\:{of}\:{land} \\ $$

Question Number 189870    Answers: 1   Comments: 0

how do i prove this using ε and δ please help lim_(x,y→(2,1)) x^2 y=4

$$ \\ $$$$\:{how}\:{do}\:{i}\:{prove}\:{this}\:{using}\:\epsilon\:{and}\:\delta \\ $$$$\:{please}\:{help} \\ $$$$\underset{{x},{y}\rightarrow\left(\mathrm{2},\mathrm{1}\right)} {\mathrm{lim}}\:{x}^{\mathrm{2}} {y}=\mathrm{4} \\ $$$$ \\ $$

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 189864    Answers: 1   Comments: 0

Question Number 189861    Answers: 1   Comments: 0

Question Number 189860    Answers: 0   Comments: 1

Question Number 189854    Answers: 0   Comments: 1

1−determiner l aire de la partie hachuree(SDEB) 2−le raport (R_1 /R_2 ) sachant que: AB=20(diametre de (C) ∡CDE=60^° ∡OAD=30°. [AI∣∣EF

$$\mathrm{1}−{determiner}\:{l}\:{aire}\:{de}\:{la} \\ $$$${partie}\:{hachuree}\left({SDEB}\right) \\ $$$$\mathrm{2}−{le}\:{raport}\:\frac{{R}_{\mathrm{1}} }{{R}_{\mathrm{2}} } \\ $$$${sachant}\:{que}:\:{AB}=\mathrm{20}\left({diametre}\:{de}\:\left({C}\right)\right. \\ $$$$\:\:\measuredangle{CDE}=\mathrm{60}^{°} \:\:\:\measuredangle{OAD}=\mathrm{30}°.\:\:\left[{AI}\mid\mid{EF}\right. \\ $$$$\:\: \\ $$

Question Number 189859    Answers: 0   Comments: 0

Question Number 189858    Answers: 1   Comments: 0

Question Number 189848    Answers: 0   Comments: 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 189843    Answers: 1   Comments: 0

Question Number 189838    Answers: 0   Comments: 0

Question Number 189826    Answers: 0   Comments: 0

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