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Question Number 137252    Answers: 1   Comments: 0

Decompose the function P(x) = ((x^4 +2x^3 +6x^2 +20x+6)/(x^3 +x^2 +x)) in partial fractions.

DecomposethefunctionP(x)=x4+2x3+6x2+20x+6x3+x2+xinpartialfractions.

Question Number 137191    Answers: 0   Comments: 0

Some useful approximations of sine function sin((π/7))=((96)/(221)) sin((π/9))=((128)/(373)) sin((π/(11)))=((32)/(113)) ... I am counting more ..thanking you!

Someusefulapproximationsofsinefunctionsin(π7)=96221sin(π9)=128373sin(π11)=32113...Iamcountingmore..thankingyou!

Question Number 136995    Answers: 0   Comments: 0

1−(((1.1.3)/(2.3.4)))(1/(1!))+(((3.3.7)/(2^2 .3^2 .4^2 )))(1/(2!))−(((5.7.10)/(2^3 .3^3 .4^3 )))(1/(3!))−....

1(1.1.32.3.4)11!+(3.3.722.32.42)12!(5.7.1023.33.43)13!....

Question Number 137016    Answers: 1   Comments: 0

Question Number 136950    Answers: 2   Comments: 0

Question Number 136930    Answers: 0   Comments: 6

Question Number 136899    Answers: 2   Comments: 0

Question Number 136892    Answers: 1   Comments: 0

∫(d^2 y/dx^2 )dy

d2ydx2dy

Question Number 136885    Answers: 1   Comments: 0

Question Number 136850    Answers: 0   Comments: 1

Σ_(n=−∞) ^∞ a^((n(n+1))/2) b^((n(n−1))/2) =1+(√((2a^2 )/π))∫_0 ^∞ e^(−t^2 /2) (((1−a(√(ab)) cosh((√(log(ab))) t))/(a^3 b−2a(√(ab )) cosh((√(log(ab))) t))))dt

n=an(n+1)2bn(n1)2=1+2a2π0et2/2(1aabcosh(log(ab)t)a3b2aabcosh(log(ab)t))dt

Question Number 136826    Answers: 0   Comments: 0

Question Number 136823    Answers: 0   Comments: 0

Question Number 136799    Answers: 0   Comments: 0

∫_0 ^1 Π_(n=1) ^∞ (1−q^n )dq

01n=1(1qn)dq

Question Number 136761    Answers: 1   Comments: 1

Question Number 136760    Answers: 1   Comments: 0

Question Number 136757    Answers: 1   Comments: 0

1+((1/2))^2 (1/(2.1!))+(((1.3)/2^2 ))^2 (1/(2^2 .2!))+(((1.3.5)/2^3 )).(1/(2^3 .3!))+....=(((Γ((1/4)))/((2π^3 )^(1/4) )))^2

1+(12)212.1!+(1.322)2122.2!+(1.3.523).123.3!+....=(Γ(14)(2π3)1/4)2

Question Number 136661    Answers: 0   Comments: 0

∫_0 ^1 (1−x)^(−(7/8)) (1+x)^(−(1/(16))) dx

01(1x)78(1+x)116dx

Question Number 136593    Answers: 1   Comments: 0

Question Number 136520    Answers: 3   Comments: 1

Question Number 136519    Answers: 1   Comments: 0

1−((1/2))^(2k) +(((1.3)/(2.4)))^(2k) −(((1.3.5)/(2.4.6)))^(2k) +(((1.3.5.7)/(2.4.6.8)))^(2k) −.... Find the general form

1(12)2k+(1.32.4)2k(1.3.52.4.6)2k+(1.3.5.72.4.6.8)2k....Findthegeneralform

Question Number 136420    Answers: 2   Comments: 0

Σ_(n=1) ^∞ ((sin(2n+1)θ)/((2n+1)^2 ))=(θ/2)log(2)+sinθ((log(sinθ))/4) _2 F_1 ((1/2),(1/2);(3/2);sin^2 θ)+((sinθ)/(16)) _2 F_1 ((1/2),(1/2),(1/2);(3/2);(3/2);sin^2 θ) Prove or disprove 0<θ<π

n=1sin(2n+1)θ(2n+1)2=θ2log(2)+sinθlog(sinθ)42F1(12,12;32;sin2θ)+sinθ162F1(12,12,12;32;32;sin2θ)Proveordisprove0<θ<π

Question Number 136105    Answers: 0   Comments: 4

Question Number 136104    Answers: 0   Comments: 0

((cos(1+(√((43)/3)))π)/1^2 )+((cos(1+(√((43)/3)))2π)/2^2 )+((cos(1+(√((43)/3)))3π)/3^2 )+...=aπ Find a

cos(1+433)π12+cos(1+433)2π22+cos(1+433)3π32+...=aπFinda

Question Number 136070    Answers: 1   Comments: 0

((sin((√2)))/1^3 )+((sin(2(√2)))/2^3 )+((sin(3(√2)))/3^3 )+...=((π^b +1)/( a(√b)))−(π/b) Find a−b

sin(2)13+sin(22)23+sin(32)33+...=πb+1abπbFindab

Question Number 136110    Answers: 0   Comments: 1

An engine is pumping water from a well 25m deep. It discharges 0.4 m^3 of water each second with a velocity of 12 ms^(−1) . Find the power of the pump given that the density of water is 1000 kg m^(−3) . take g = 10 ms^(−2)

Anengineispumpingwaterfromawell25mdeep.Itdischarges0.4m3ofwatereachsecondwithavelocityof12ms1.Findthepowerofthepumpgiventhatthedensityofwateris1000kgm3.takeg=10ms2

Question Number 135976    Answers: 1   Comments: 0

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