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Question Number 111850    Answers: 4   Comments: 0

If ∣z∣ = 3 , what is the maximum and minimum value of ∣z−1+i(√3) ∣ ?

Ifz=3,whatisthemaximumandminimumvalueofz1+i3?

Question Number 112531    Answers: 0   Comments: 4

A rectangular cardboard is 8cm long and 6cm wide. What is the least number of beads you can arrange on the board such that there are at least two of the beads that are less than (√(10))cm apart.

Arectangularcardboardis8cmlongand6cmwide.Whatistheleastnumberofbeadsyoucanarrangeontheboardsuchthatthereareatleasttwoofthebeadsthatarelessthan10cmapart.

Question Number 111831    Answers: 1   Comments: 0

Let N be the greatest multiple of 36 all of whose digits are even and no two of whose digits are the same. Find the remainder when N is divided by 1000.

LetNbethegreatestmultipleof36allofwhosedigitsareevenandnotwoofwhosedigitsarethesame.FindtheremainderwhenNisdividedby1000.

Question Number 112535    Answers: 1   Comments: 5

Let K be the product of all factors (b−a) (not necessarily distinct) where a and b are integers satisfying 1≤a≤b≤10. Find the greatest integer n such that 2^n divides K.

LetKbetheproductofallfactors(ba)(notnecessarilydistinct)whereaandbareintegerssatisfying1ab10.Findthegreatestintegernsuchthat2ndividesK.

Question Number 111818    Answers: 3   Comments: 0

(√(bemath )) (1)∫ ((cos x)/(2−cos x)) dx (2) f(x) = ∣x^3 ∣ ⇒ f ′(x) ?

bemath(1)cosx2cosxdx(2)f(x)=x3f(x)?

Question Number 111813    Answers: 0   Comments: 3

if ψ(x)=(x/(1−x))+(x^3 /(1−x^2 ))+(x^5 /(1−x^5 ))+(x^7 /(1−x^7 )) show that ∫_0 ^1 [ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )]((ln((1/x)))/x)=((25832π^2 )/(1575)) soltion let the generating series form of ψ(x)=(x/(1−x))+(x^3 /(1−x^2 ))+(x^5 /(1−x^5 ))+(x^7 /(1−x^7 )) be ψ(x)=Σ_(i=0) ^∞ (x^(2i+1) /(1−x^(2i+1) )).........(1) from Ω=∫_0 ^1 ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )((ln((1/x)))/x)dx Ω=−∫_0 ^1 [ψ(x)+ψ(x^(1/3) )+ψ(x^(1/5) )+ψ(x^(1/7) )]((lnx)/x)dx there the series form be Ω=∫_0 ^1 [Σ_(n=0) ^∞ ψ(x^(1/(2n+1)) )((lnx)/x)]dx.........(2) putting equation (1) into (2) Ω=−∫_0 ^1 Σ_(n=0) ^∞ Σ_(i=0) ^∞ [(((x^(2i+1) .x^(1/(2n+1)) )/(1−x^(2i+1) .x^(1/(2n+1)) )))((lnx)/x)]dx Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 [((x^((2i+1)/(2n+1)) /(1−x^((2i+1)/(2n+1)) )))((lnx)/x)]dx let y=((2i+1)/(2n+1)) Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 [((x^y /(1−x^y )))((lnx)/x)]dx=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 ((x^(y−1) /(1−x^y )))lnxdx let the series form of (1/(1−x^y ))=Σ_(m=0) ^∞ x^(my) Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 (Σ_(m=0) ^∞ x^(my) .x^(y−1) lnx)dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ∫_0 ^1 (x^(my) .x^(y−1) .x^a )dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ∫_0 ^1 x^(my+y+a−1) dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ((1/(my+y+a))) Ω^′ (0)=Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ [(1/((my+y)^2 ))]=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )Σ_(m=0) ^∞ (1/((1+m)^2 )) let z=1+m at m=0 ,z=1 Ω(0)=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )Σ_(z=1) ^∞ (1/z^2 )=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )ζ(2)=(π^2 /6)(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 ) but y=((2i+1)/(2n+1)) Ω(0)=(π^2 /6)(Σ_(n=0) ^∞ Σ_(i=0) ^∞ (1/((((2i+1)/(2n+1)))^2 )))=(π^2 /6)Σ_(n=0) ^∞ (2n+1)^2 Σ_(i=0) ^∞ (1/((2i+1)^2 )) Ω=(π^2 /6)[1^2 +3^2 +5^2 +7^2 ]×[(1/1^2 )+(1/3^2 )+(1/5^2 )+(1/7^2 )] Ω=(π^2 /6)(1+9+25+49)×[(1/1)+(1/9)+(1/(25))+(1/(49))] Ω=(π^2 /6)•(84)•((12916)/(11025))=((25832π^2 )/(1575)) ∵∫_0 ^1 [ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )]((ln((1/x)))/x)=((25832π^2 )/(1575)) by mathew monday(05/09/2020)

ifψ(x)=x1x+x31x2+x51x5+x71x7showthat01[ψ(x)+ψ(x3)+ψ(x5)+ψ(x7)]ln(1x)x=25832π21575soltionletthegeneratingseriesformofψ(x)=x1x+x31x2+x51x5+x71x7beψ(x)=i=0x2i+11x2i+1.........(1)fromΩ=01ψ(x)+ψ(x3)+ψ(x5)+ψ(x7)ln(1x)xdxΩ=01[ψ(x)+ψ(x13)+ψ(x15)+ψ(x17)]lnxxdxtheretheseriesformbeΩ=01[n=0ψ(x12n+1)lnxx]dx.........(2)puttingequation(1)into(2)Ω=01n=0i=0[(x2i+1.x12n+11x2i+1.x12n+1)lnxx]dxΩ=n=0i=001[(x2i+12n+11x2i+12n+1)lnxx]dxlety=2i+12n+1Ω=n=0i=001[(xy1xy)lnxx]dx=n=0i=001(xy11xy)lnxdxlettheseriesformof11xy=m=0xmyΩ=n=0i=001(m=0xmy.xy1lnx)dxaa=0Ω(a)=n=0i=0m=001(xmy.xy1.xa)dxaa=0Ω(a)=n=0i=0m=001xmy+y+a1dxaa=0Ω(a)=n=0i=0m=0(1my+y+a)Ω(0)=n=0i=0m=0[1(my+y)2]=(n=0i=0)1y2m=01(1+m)2letz=1+matm=0,z=1Ω(0)=(n=0i=0)1y2z=11z2=(n=0i=0)1y2ζ(2)=π26(n=0i=0)1y2buty=2i+12n+1Ω(0)=π26(n=0i=01(2i+12n+1)2)=π26n=0(2n+1)2i=01(2i+1)2Ω=π26[12+32+52+72]×[112+132+152+172]Ω=π26(1+9+25+49)×[11+19+125+149]Ω=π26(84)1291611025=25832π2157501[ψ(x)+ψ(x3)+ψ(x5)+ψ(x7)]ln(1x)x=25832π21575bymathewmonday(05/09/2020)

Question Number 111799    Answers: 3   Comments: 4

Question Number 111788    Answers: 1   Comments: 0

Question Number 111781    Answers: 0   Comments: 3

Question Number 111774    Answers: 1   Comments: 0

Question Number 111772    Answers: 1   Comments: 0

calculate lim_(n→+∞) Π_(k=1) ^n (1+((√(k(n−k)))/n^2 ))

calculatelimn+k=1n(1+k(nk)n2)

Question Number 111771    Answers: 0   Comments: 0

calculate lim_(n→+∞) a_n ∫_0 ^1 x^(2n) sin(((πx)/2))dx with a_n =Σ_(k=1) ^n sin(((πk)/(2n)))

calculatelimn+an01x2nsin(πx2)dxwithan=k=1nsin(πk2n)

Question Number 111770    Answers: 0   Comments: 0

f function continue on [0,1] find lim_(n→+∞) (1/n)Σ_(k=0) ^n (n−k)∫_(k/n) ^((k+1)/n) f(x)dx

ffunctioncontinueon[0,1]findlimn+1nk=0n(nk)knk+1nf(x)dx

Question Number 111769    Answers: 0   Comments: 0

let u_n =Σ_(k=1) ^n (1/(√((k+n)(k+n+1)))) detremine lim_(n→+∞) u_n

letun=k=1n1(k+n)(k+n+1)detreminelimn+un

Question Number 111768    Answers: 0   Comments: 0

explicite f(x) =∫_0 ^(2π) ln(x^2 −2xcosθ +1)dθ (x≠+^− 1)

explicitef(x)=02πln(x22xcosθ+1)dθ(x+1)

Question Number 111766    Answers: 2   Comments: 0

calculate lim_(n→+∞) Σ_(k=1) ^n ln((n/(n+k)))^(1/n)

calculatelimn+k=1nln(nn+k)1n

Question Number 111762    Answers: 1   Comments: 0

caoculate ∫_0 ^(π/4) ln(1+2tanx)dx

caoculate0π4ln(1+2tanx)dx

Question Number 111760    Answers: 0   Comments: 0

find lim_(x→1^+ ) ∫_x ^x^2 ((ln(t))/((t−1)^2 ))dx

findlimx1+xx2ln(t)(t1)2dx

Question Number 111756    Answers: 0   Comments: 0

find I_n =∫_0 ^(π/4) (du/(cos^n u))

findIn=0π4ducosnu

Question Number 111755    Answers: 2   Comments: 0

calculate lim_(n→+∞) Σ_(k=1) ^n (√((n−k)/(n^3 +n^2 k)))

calculatelimn+k=1nnkn3+n2k

Question Number 111754    Answers: 2   Comments: 0

calculate lim_(n→+∞) Σ_(k=1) ^(2n) (k/(k^2 +n^2 ))

calculatelimn+k=12nkk2+n2

Question Number 111887    Answers: 0   Comments: 0

Question Number 111882    Answers: 1   Comments: 0

(√(bemath)) (1)lim_(x→∞) [(x^2 /(x+1)) − (x^2 /(x+3)) ] ? (2) prove that n^2 ≤ 2^n for ∀n∈N by mathematical induction

bemath(1)limx[x2x+1x2x+3]?(2)provethatn22nfornNbymathematicalinduction

Question Number 111745    Answers: 0   Comments: 6

Question Number 111826    Answers: 0   Comments: 2

There are two cards; one is yellow on both sides and the other one is yellow in one side and red on the other. The cards have the same probability ((1/2)) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is yellow, then the probability that the under side is also yellow is.

Therearetwocards;oneisyellowonbothsidesandtheotheroneisyellowinonesideandredontheother.Thecardshavethesameprobability(12)ofbeingchosen,andoneischosenandplacedonthetable.Iftheuppersideofthecardonthetableisyellow,thentheprobabilitythattheundersideisalsoyellowis.

Question Number 111734    Answers: 1   Comments: 0

A chord which is a perpendicular bisector of radius of length 18cm in a circle, has length.

Achordwhichisaperpendicularbisectorofradiusoflength18cminacircle,haslength.

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