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Question Number 192606    Answers: 0   Comments: 1

proof that 0^0 =1 without limit!

proofthat00=1withoutlimit!

Question Number 192604    Answers: 0   Comments: 1

many people say that the 0^(0 ) is an uninfinity ones of them say that 0^0 is infinity and equal to 1! what do you think wich ones of them say right?

manypeoplesaythatthe00isanuninfinityonesofthemsaythat00isinfinityandequalto1!whatdoyouthinkwichonesofthemsayright?

Question Number 192597    Answers: 1   Comments: 0

a_(1 ) , a_2 , a_(3 ) , .... , a_n is a sequence satifies that a_(n+2) =a_(n+1) −a_n for n ≥ 1. suppose the sum of the first 999 terms = 1003 and the sum of the first 1003 terms = −999 find the sum of the first 2002 terms.

a1,a2,a3,....,anisasequencesatifiesthatan+2=an+1anforn1.supposethesumofthefirst999terms=1003andthesumofthefirst1003terms=999findthesumofthefirst2002terms.

Question Number 192596    Answers: 2   Comments: 0

Question Number 192594    Answers: 0   Comments: 0

Question Number 192593    Answers: 0   Comments: 0

Question Number 192590    Answers: 1   Comments: 0

Σ_(n=1) ^∞ Σ_(m=1) ^∞ [((m^2 n)/(3^m (3^n .m+3^m .n)))]=λ find λ.

n=1m=1[m2n3m(3n.m+3m.n)]=λfindλ.

Question Number 192584    Answers: 0   Comments: 0

Question Number 192582    Answers: 1   Comments: 5

Question Number 192580    Answers: 0   Comments: 0

This might be helpful: How to easily calculate z_1 ^z_2 with z_1 , z_2 ∈C: Transform z_1 =re^(iθ) and z_2 =a+bi ⇒ z_1 ^z_2 =(re^(iθ) )^(a+bi) z_1 ^z_2 =r^(a+bi) e^(−bθ+iaθ) z_1 ^z_2 =r^a e^(−bθ) r^(bi) e^(iaθ) z_1 ^z_2 =(r^a /e^(bθ) )e^(i(aθ+bln r))

Thismightbehelpful:Howtoeasilycalculatez1z2withz1,z2C:Transformz1=reiθandz2=a+biz1z2=(reiθ)a+biz1z2=ra+biebθ+iaθz1z2=raebθrbieiaθz1z2=raebθei(aθ+blnr)

Question Number 192560    Answers: 1   Comments: 3

Question Number 192572    Answers: 0   Comments: 0

Let ABCD be a rectangle having an area of 290. Let E be on BC such that BE : BC = 3 : 2. Let F be on CD such that CF : FD = 3 : 1. If G is the intersection of AE and BF, compute the area of △BEG.

LetABCDbearectanglehavinganareaof290.LetEbeonBCsuchthatBE:BC=3:2.LetFbeonCDsuchthatCF:FD=3:1.IfGistheintersectionofAEandBF,computetheareaofBEG.

Question Number 192570    Answers: 2   Comments: 0

Question Number 192573    Answers: 0   Comments: 0

solve; lim_(x→0) x^2 tan(((sinπx)/(2x))) solution let L=lim_(x→0) x^2 tan(((sinπx)/(2x))) since sinx∼x−(x^3 /6) L=lim_(x→0) x^2 tan(((πx)/(2x))−((π^3 x^3 )/(12x))) L=lim_(x→0) x^2 tan((π/2)−((π^3 x^2 )/(12))) since tan((π/2)−x)=(1/(tanx)) L=lim_(x→0) (x^2 /(tan(((π^3 x^2 )/(12))))) L=lim_(x→0) (((π^3 x^2 )/(12))/(tan(((π^3 x^2 )/(12))))) ((12)/π^3 ) L=((12)/π^3 )lim_(x→0) (((π^3 x^2 )/(12))/(tan(((π^3 x^2 )/(12))))) since lim_(x→0) (x/(tanx))=1 L=((12)/π^3 ) ∙1=((12)/π^3 ) solved by HY a.k.a senestro

solve;limx0x2tan(sinπx2x)solutionletL=limx0x2tan(sinπx2x)sincesinxxx36L=limx0x2tan(πx2xπ3x312x)L=limx0x2tan(π2π3x212)sincetan(π2x)=1tanxL=limx0x2tan(π3x212)L=limx0π3x212tan(π3x212)12π3L=12π3limx0π3x212tan(π3x212)sincelimx0xtanx=1L=12π31=12π3solvedbyHYa.k.asenestro

Question Number 192550    Answers: 1   Comments: 0

how can find the sum Σ_(i=1) ^r (2v_i +1) ?

howcanfindthesumri=1(2vi+1)?

Question Number 192549    Answers: 2   Comments: 0

lim_(x→0) x^2 tan (((sin πx)/(2x))) =?

limx0x2tan(sinπx2x)=?

Question Number 192544    Answers: 0   Comments: 1

If f(t) = 2(e^t − 1) a) Does f(t) exists for all n ? b) If it exist, does it converge ? c) If the sequence converge, does the limit converge ? d) Is the solution uniques ?

Iff(t)=2(et1)a)Doesf(t)existsforalln?b)Ifitexist,doesitconverge?c)Ifthesequenceconverge,doesthelimitconverge?d)Isthesolutionuniques?

Question Number 192543    Answers: 1   Comments: 0

Question Number 192542    Answers: 2   Comments: 0

Question Number 192540    Answers: 0   Comments: 1

Question Number 192541    Answers: 0   Comments: 0

Question Number 192537    Answers: 1   Comments: 0

{ ((tan (α+2β)=(√(1+2k)))),((tan^2 (α+β){1+k tan^2 β}=k+tan^2 β)) :} Find cot 2β .

{tan(α+2β)=1+2ktan2(α+β){1+ktan2β}=k+tan2βFindcot2β.

Question Number 192536    Answers: 1   Comments: 0

R=((3sin 5°+4cos 5°−5cos 58°+35(√2) cos 13°)/(cos 5°))=?

R=3sin5°+4cos5°5cos58°+352cos13°cos5°=?

Question Number 192545    Answers: 3   Comments: 0

Question Number 192514    Answers: 1   Comments: 2

Question Number 192508    Answers: 1   Comments: 1

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