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Question Number 138154    Answers: 1   Comments: 0

Question Number 138153    Answers: 1   Comments: 0

lim_(x→0^+ ) ((x−∣tanx∣)/(∣sinx∣−x))=?

limx0+xtanxsinxx=?

Question Number 138150    Answers: 2   Comments: 0

Given (x+(√(x^2 +1)))(y+(√(y^2 +4)))=9 find the value of x(√(y^2 +4)) + y(√(x^2 +1)) .

Given(x+x2+1)(y+y2+4)=9findthevalueofxy2+4+yx2+1.

Question Number 138145    Answers: 4   Comments: 0

...........advanced ... ... ... calculus......... find the value of:: Θ=Σ_(n=1) ^∞ (((−1)^n H_(2n) )/n)=???

...........advanced.........calculus.........findthevalueof::Θ=n=1(1)nH2nn=???

Question Number 138140    Answers: 4   Comments: 0

lim_(x→0) ((cos (sin x)−cos x)/x^4 )=?

limx0cos(sinx)cosxx4=?

Question Number 138135    Answers: 1   Comments: 0

(x^2 +x^3 )=4 find x.

(x2+x3)=4findx.

Question Number 138134    Answers: 0   Comments: 0

calculate Σ_(n=1) ^∞ (((−1)^n )/(n^3 (2n+1)^4 ))

calculaten=1(1)nn3(2n+1)4

Question Number 138133    Answers: 0   Comments: 0

let A = (((1 −1)),((2 3)) ) 1)calculate e^A and e^(tA) 2)find cosA ,sinA 3) find log(1+A)

letA=(1123)1)calculateeAandetA2)findcosA,sinA3)findlog(1+A)

Question Number 138132    Answers: 1   Comments: 0

calculate ∫_0 ^∞ ((logx)/(x^2 −x+2))dx

calculate0logxx2x+2dx

Question Number 138131    Answers: 1   Comments: 0

calculate ∫_0 ^∞ (dx/((2x+1)(x^2 −x+3)^2 ))

calculate0dx(2x+1)(x2x+3)2

Question Number 138129    Answers: 0   Comments: 0

calculate ∫_0 ^1 ((ln(2+x^2 ))/(x^2 +3))dx

calculate01ln(2+x2)x2+3dx

Question Number 138128    Answers: 1   Comments: 0

calculate ∫ (dx/( (√(2−x^2 ))+(√(3+x^2 ))))

calculatedx2x2+3+x2

Question Number 138123    Answers: 1   Comments: 0

Question Number 138120    Answers: 2   Comments: 1

Question Number 138114    Answers: 1   Comments: 2

...... calculus.....(III)...... evaluate:: 𝛗=^(???) ∫_0 ^( (π/2)) ∫_0 ^( x) ((cos(y))/( (√(((π/2)−x)((π/2)−y)))))dydx

......calculus.....(III)......evaluate::ϕ=???0π20xcos(y)(π2x)(π2y)dydx

Question Number 138112    Answers: 2   Comments: 0

prove that ∫_0 ^( (π/2)) (sin^(2k) (x)dx)=(((2k)!2^(−2k−1) )/((k!)^2 ))

provethat0π2(sin2k(x)dx)=(2k)!22k1(k!)2

Question Number 138099    Answers: 0   Comments: 5

If sin^4 α+cos^4 β = 4sin α cos β , 0≤α,β ≤ (π/2) , then sin α+cos β equal to …

Ifsin4α+cos4β=4sinαcosβ,0α,βπ2,thensinα+cosβequalto

Question Number 138096    Answers: 1   Comments: 1

solve 2^x +4^x =8^x

solve2x+4x=8x

Question Number 138091    Answers: 2   Comments: 0

If 3^((log _3 7)^x ) = 7^((log _7 3)^x ) , then the value of x will be …

If3(log37)x=7(log73)x,thenthevalueofxwillbe

Question Number 138089    Answers: 1   Comments: 1

Question Number 138085    Answers: 1   Comments: 0

Given a curve y = (1/(x^2 +1)). Find the equation of tangent line with have slope of tangent minimum .

Givenacurvey=1x2+1.Findtheequationoftangentlinewithhaveslopeoftangentminimum.

Question Number 138086    Answers: 0   Comments: 0

∫_0 ^(π/2) ln(((ln^2 (sinθ))/(π^2 +ln^2 (sinθ)))) ((ln(cosθ))/(tanθ))dθ

0π2ln(ln2(sinθ)π2+ln2(sinθ))ln(cosθ)tanθdθ

Question Number 138077    Answers: 2   Comments: 0

∫_( 0) ^( (π/2)) (((cotx))^(1/3) +(1/( ((cotx))^(1/3) )))^2 dx

0π2(cotx3+1cotx3)2dx

Question Number 138075    Answers: 0   Comments: 0

Use the inner product ⟨f,g⟩=∫10f(x)g(x)dx in the vector space C0[0,1] of continuous functions on the domain [0,1] to find the orthogonal projection of f(x)=3x2−2 onto the subspace V spanned by g(x)=x and h(x)=1. (Caution: x and 1 do not form an orthogonal basis of V.) projV(f)=  .

Use the inner product ⟨f,g⟩=∫10f(x)g(x)dx in the vector space C0[0,1] of continuous functions on the domain [0,1] to find the orthogonal projection of f(x)=3x2−2 onto the subspace V spanned by g(x)=x and h(x)=1. (Caution: x and 1 do not form an orthogonal basis of V.) projV(f)=  .

Question Number 138074    Answers: 0   Comments: 0

 Let W be the set of all vectors ⎡⎣⎢xyx+y⎤⎦⎥ with x and y real. Find a basis of W⊥. { ⎡⎣⎢⎢⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎥⎥ }

 Let W be the set of all vectors ⎡⎣⎢xyx+y⎤⎦⎥ with x and y real. Find a basis of W⊥. { ⎡⎣⎢⎢⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎥⎥ }

Question Number 138073    Answers: 0   Comments: 0

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