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Question Number 176684    Answers: 1   Comments: 2

Question Number 176679    Answers: 2   Comments: 0

Eeasy integral.... 𝛀 = ∫_(−∫_0 ^( ∞) e^( −x^( 2) ) dx) ^( ∫_0 ^( ∞) e^( −x^( 2) ) dx) sin^( 2) (t).ln^( 3) ( t + (√(1+t^( 2) )))dt −−−m.n−−−

Eeasyintegral....Ω=∫−∫0∞e−x2dx∫0∞e−x2dxsin2(t).ln3(t+1+t2)dt−−−m.n−−−

Question Number 176676    Answers: 3   Comments: 0

If x^3 +(1/x^3 )=1, prove that x^5 +(1/x^5 )=−(x^4 +(1/x^4 ))

Ifx3+1x3=1,provethatx5+1x5=−(x4+1x4)

Question Number 176672    Answers: 1   Comments: 0

Question Number 176673    Answers: 1   Comments: 0

Question Number 176668    Answers: 3   Comments: 0

Question Number 176662    Answers: 3   Comments: 0

sequence V_(n+1) −V_n =n+3^n . Find V_n .

sequenceVn+1−Vn=n+3n.FindVn.

Question Number 176660    Answers: 1   Comments: 1

Question Number 176652    Answers: 1   Comments: 0

f(x,y)= { ((e^(1/(r^2 −1)) if r<1, where r=∥(x,y)∥)),((0 if r≥1)) :} show that f(x,y) is continuous in R^2

f(x,y)={e1r2−1ifr<1,wherer=∥(x,y)∥0ifr⩾1showthatf(x,y)iscontinuousinR2

Question Number 176648    Answers: 2   Comments: 1

∫ (dx/(a+bcosx)) ∫ (dx/(a−bsinx))

∫dxa+bcosx∫dxa−bsinx

Question Number 176643    Answers: 0   Comments: 2

Question Number 176638    Answers: 1   Comments: 0

lim_(x→0) ((sin^2 (x)−sin (x^2 ))/(x^2 (cos^2 (x)−cos (x^2 ))))=?

limx→0sin2(x)−sin(x2)x2(cos2(x)−cos(x2))=?

Question Number 176637    Answers: 0   Comments: 0

Question Number 176636    Answers: 2   Comments: 2

{ ((p^3 +q^3 =r^2 )),((p^3 +r^3 =q^2 )),((q^3 +r^3 =p^2 )) :} ⇒20pqr =?

{p3+q3=r2p3+r3=q2q3+r3=p2⇒20pqr=?

Question Number 176610    Answers: 1   Comments: 0

Question Number 176607    Answers: 1   Comments: 0

Question Number 176603    Answers: 2   Comments: 6

look the anser

looktheanser

Question Number 176595    Answers: 2   Comments: 0

Question Number 176594    Answers: 0   Comments: 1

𝛗=∫_0 ^( 1) (( ( tanh^( −1) (x))^2 )/((1+x )^( 2) )) dx = ? ≺ solution ≻ note : tanh^( −1) (x)=− (1/2) ln(((1−x)/(1+x))) 𝛗= (1/4)∫_0 ^( 1) (( ln^( 2) (((1−x)/(1+x)) ))/((1+x )^( 2) )) dx =^(((1−x)/(1+x)) = t) (1/8)∫_0 ^( 1) ln^( 2) (t )dt =(1/8_ ) { [t.ln^( 2) (t)]_0 ^( 1) −2∫_0 ^( 1) ln(t)dt} =− (1/4) ∫_0 ^( 1) ln(t)dt= (1/4) ◂ m.n ▶

ϕ=∫01(tanh−1(x))2(1+x)2dx=?≺solution≻note:tanh−1(x)=−12ln(1−x1+x)ϕ=14∫01ln2(1−x1+x)(1+x)2dx=1−x1+x=t18∫01ln2(t)dt=18{[t.ln2(t)]01−2∫01ln(t)dt}=−14∫01ln(t)dt=14◂m.n▸

Question Number 176592    Answers: 2   Comments: 0

Solve the equation: 2^x + 3^x − 4^x + 6^x − 9^x = 1

Solvetheequation:2x+3x−4x+6x−9x=1

Question Number 176589    Answers: 1   Comments: 0

Question Number 176598    Answers: 1   Comments: 0

x^3 +(1/x^3 )=1 (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=? Q#176387 reposted for a new answer.

x3+1x3=1(x5+1x5)3−1x5+1x5=?You can't use 'macro parameter character #' in math mode

Question Number 176581    Answers: 1   Comments: 0

Given { ((sin a+sin b=((√2)/2))),((cos a+cos b=((√6)/2))) :} for a,b real numbers. Evaluate sin (a+b). (A)((√3)/2) (D) −((√3)/2) (B) (2/( (√3))) (E)−(2/( (√3))) (C) ((√3)/4)

Given{sina+sinb=22cosa+cosb=62fora,brealnumbers.Evaluatesin(a+b).(A)32(D)−32(B)23(E)−23(C)34

Question Number 176580    Answers: 1   Comments: 0

4^(x^2 −2x+2) −2^(x^2 −2x+3) +2=2^(x^2 −2x+2) x=?

4x2−2x+2−2x2−2x+3+2=2x2−2x+2x=?

Question Number 176571    Answers: 1   Comments: 0

lim_(x→1) ((lnx)/(1+lnx−1))=?

limx→1lnx1+lnx−1=?

Question Number 176570    Answers: 2   Comments: 0

(1) ∫^(π/2) _(π/3) ((1+sinx)/(cosx)) dx=?

(1)∫π3π21+sinxcosxdx=?

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