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Question Number 137588 Answers: 1 Comments: 0

For a positive number n , let f(n) be the value of f(n)=((4n+(√(4n^2 −1)))/( (√(2n+1)) +(√(2n−1)))) calculate f(1)+f(2)+f(3)+...+f(40).

$${For}\:{a}\:{positive}\:{number}\:{n}\:,\:{let} \\ $$$${f}\left({n}\right)\:{be}\:{the}\:{value}\:{of}\: \\ $$$${f}\left({n}\right)=\frac{\mathrm{4}{n}+\sqrt{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{2}{n}+\mathrm{1}}\:+\sqrt{\mathrm{2}{n}−\mathrm{1}}} \\ $$$${calculate}\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+...+{f}\left(\mathrm{40}\right). \\ $$

Question Number 137585 Answers: 2 Comments: 0

Find the cube of the number N= (√(7(√(3(√(7(√(3(√(7(√(3(√(7(√(3...))))))))))))))))

$${Find}\:{the}\:{cube}\:{of}\:{the}\:{number}\: \\ $$$${N}=\:\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}...}}}}}}}} \\ $$

Question Number 137584 Answers: 1 Comments: 0

Given { ((a_(2n) = a_n .a_2 +1)),((a_(2n+1) = a_n .a_2 −2 )) :} and { ((a_7 = 2)),((0<a_1 <1)) :}. Find a_(25) =?

$${Given}\:\begin{cases}{{a}_{\mathrm{2}{n}} \:=\:{a}_{{n}} .{a}_{\mathrm{2}} \:+\mathrm{1}}\\{{a}_{\mathrm{2}{n}+\mathrm{1}} \:=\:{a}_{{n}} .{a}_{\mathrm{2}} \:−\mathrm{2}\:}\end{cases}\:{and} \\ $$$$\:\begin{cases}{{a}_{\mathrm{7}} \:=\:\mathrm{2}}\\{\mathrm{0}<{a}_{\mathrm{1}} <\mathrm{1}}\end{cases}.\:{Find}\:{a}_{\mathrm{25}} \:=? \\ $$$$ \\ $$

Question Number 137582 Answers: 1 Comments: 0

A=(((1/3)((2)^(1/3) −1)((2)^(1/3) +1)^3 ))^(1/3)

$$\mathrm{A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 137579 Answers: 1 Comments: 0

(−1)×(1/(π.i)) =?

$$\left(−\mathrm{1}\right)×\frac{\mathrm{1}}{\pi.{i}}\:=?\: \\ $$

Question Number 137575 Answers: 0 Comments: 0

Question Number 137574 Answers: 0 Comments: 0

Question Number 137568 Answers: 0 Comments: 2

Question Number 137563 Answers: 2 Comments: 0

let ((x^2 +y^2 )/(x^2 −y^2 ))+((x^2 −y^2 )/(x^2 +y^2 ))=k find the value of ((x^8 +y^8 )/(x^8 −y^8 ))+((x^8 −y^8 )/(x^8 +y^8 )) in terms of k

$${let}\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={k} \\ $$$${find}\:{the}\:{value}\:{of}\:\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }+\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} } \\ $$$${in}\:{terms}\:{of}\:{k} \\ $$

Question Number 137559 Answers: 0 Comments: 2

What is the volume of tetrahedron ABCD, whose vertices have the coordinates A (2, 3, 6), B (3, 2, 2), C (3, 4, 7) and D (5, 1, 8). Find the lateral surface area of the tetrahedron and find the volume of the tetrahedron?

$$ \\ $$What is the volume of tetrahedron ABCD, whose vertices have the coordinates A (2, 3, 6), B (3, 2, 2), C (3, 4, 7) and D (5, 1, 8). Find the lateral surface area of the tetrahedron and find the volume of the tetrahedron?

Question Number 137558 Answers: 1 Comments: 0

(x+y)dx + (x+y^2 )dy = 0

$$\left({x}+{y}\right){dx}\:+\:\left({x}+{y}^{\mathrm{2}} \right){dy}\:=\:\mathrm{0}\: \\ $$

Question Number 137556 Answers: 0 Comments: 0

Question Number 137547 Answers: 0 Comments: 1

Question Number 137545 Answers: 0 Comments: 0

Question Number 137538 Answers: 0 Comments: 0

calculate ∫_(−∞) ^(+∞) ((2x+1)/((x^4 −2x^2 +3)^2 ))

$${calculate}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} } \\ $$

Question Number 137536 Answers: 1 Comments: 0

calculte ∫_(−∞) ^∞ ((sin(πx^2 ))/((x^2 +2x+2)^2 ))dx

$${calculte}\:\int_{−\infty} ^{\infty} \:\frac{{sin}\left(\pi{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 137535 Answers: 0 Comments: 0

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

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{3}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

Question Number 137530 Answers: 1 Comments: 0

If sin^(−1) (sin α+sin β)+sin^(−1) (sin α−sin β)=(π/2) find the value of sin^2 α+sin^2 β. [(1/2)]

$$\mathrm{If}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\right)+\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\beta\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{sin}^{\mathrm{2}} \beta.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$

Question Number 137528 Answers: 3 Comments: 1

Question Number 137525 Answers: 0 Comments: 0

Question Number 137523 Answers: 1 Comments: 0

L=lim_(x→0) ((x!−1)/x)

$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}!−\mathrm{1}}{\mathrm{x}} \\ $$

Question Number 137518 Answers: 1 Comments: 0

Ω = ∫_0 ^( ∞) ((u^2 +2)/(u^4 +2u^2 +2))du

$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{u}^{\mathrm{2}} +\mathrm{2}}{\mathrm{u}^{\mathrm{4}} +\mathrm{2u}^{\mathrm{2}} +\mathrm{2}}\mathrm{du} \\ $$

Question Number 137509 Answers: 2 Comments: 1

∫ ((sin ((√x) )+cos ((√x) ))/( (√x) sin (2(√x) ))) dx

$$\int\:\frac{\mathrm{sin}\:\left(\sqrt{{x}}\:\right)+\mathrm{cos}\:\left(\sqrt{{x}}\:\right)}{\:\sqrt{{x}}\:\mathrm{sin}\:\left(\mathrm{2}\sqrt{{x}}\:\right)}\:{dx}\: \\ $$

Question Number 137508 Answers: 2 Comments: 0

∫ ((sin ((1/x)))/x^3 ) dx =?

$$\int\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{3}} }\:{dx}\:=? \\ $$

Question Number 137501 Answers: 1 Comments: 0

....advanced .... calculus.... prove that:: 𝛗=∫_0 ^( 1) ((ln(x).ln(1−x))/(1+x))dx=((13)/8)ζ(3)−(π^2 /4)ln(2)....

$$\:\:\:\:\:\:\:....{advanced}\:....\:{calculus}.... \\ $$$$\:\:{prove}\:{that}:: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right).{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{dx}=\frac{\mathrm{13}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\left(\mathrm{2}\right).... \\ $$

Question Number 137482 Answers: 2 Comments: 0

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