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Question Number 221754 Answers: 1 Comments: 0

((81))^(1/((64))^(1/((27))^(1/( 3^4^0^4^3 )) ) ) )^((√4))

$$\left.\sqrt[{\sqrt[{\sqrt[{\:\mathrm{3}^{\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } } }]{\mathrm{27}}}]{\mathrm{64}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$

Question Number 221733 Answers: 0 Comments: 19

Question Number 221721 Answers: 2 Comments: 0

Question Number 221707 Answers: 1 Comments: 13

Question Number 221697 Answers: 2 Comments: 0

Is (√i) an imaginary number (i=(√(−1))) answer with logic

$${Is}\:\sqrt{{i}}\:{an}\:{imaginary}\:{number}\:\left({i}=\sqrt{−\mathrm{1}}\right)\:{answer}\:{with}\:{logic} \\ $$

Question Number 221686 Answers: 3 Comments: 2

Question Number 221668 Answers: 0 Comments: 0

Question Number 221663 Answers: 0 Comments: 3

Prove:∫_0 ^1 Π_(k=1) ^∞ (1−x^k )dx=((4π(√3))/( (√(23))))∙((sinh(((√(23))π)/6))/(2 cosh(((√(23))π)/3)−1))

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{x}^{{k}} \right){dx}=\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{23}}}\centerdot\frac{\mathrm{sinh}\frac{\sqrt{\mathrm{23}}\pi}{\mathrm{6}}}{\mathrm{2}\:\mathrm{cosh}\frac{\sqrt{\mathrm{23}}\pi}{\mathrm{3}}−\mathrm{1}} \\ $$

Question Number 221661 Answers: 1 Comments: 1

Question Number 221647 Answers: 0 Comments: 1

solve for x. x^1 + x^2 + x^3 = 4096

$${solve}\:{for}\:{x}. \\ $$$${x}^{\mathrm{1}} \:+\:{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{3}} \:\:=\:\:\mathrm{4096} \\ $$

Question Number 221638 Answers: 1 Comments: 0

Question Number 221637 Answers: 1 Comments: 0

Question Number 221626 Answers: 3 Comments: 0

Question Number 221620 Answers: 1 Comments: 0

Solve for x ((7x))^(1/3) =(√x)[x≠0]

$${Solve}\:{for}\:{x} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{7}{x}}=\sqrt{{x}}\left[{x}\neq\mathrm{0}\right] \\ $$

Question Number 221618 Answers: 3 Comments: 0

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

$${solve}\:{for}\:{x} \\ $$$$\mathrm{2}^{{x}} +\mathrm{4}^{{x}} =\mathrm{8}^{{x}} \\ $$

Question Number 221669 Answers: 2 Comments: 3

Question Number 221601 Answers: 2 Comments: 0

∫ ((sin 2x)/(1 + sin 3x)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{{sin}\:\mathrm{2}{x}}{\mathrm{1}\:+\:{sin}\:\mathrm{3}{x}}\:{dx} \\ $$$$ \\ $$

Question Number 221592 Answers: 2 Comments: 0

Question Number 221588 Answers: 1 Comments: 0

∫ ((8t − 8t^( 3) )/(t^( 6) + 6t^5 + 3t^( 4) − 20t^3 + 3t^2 + 6t + 1)) dt

$$ \\ $$$$\:\:\:\:\int\:\frac{\mathrm{8}{t}\:−\:\mathrm{8}{t}^{\:\mathrm{3}} }{{t}^{\:\mathrm{6}} \:+\:\mathrm{6}{t}^{\mathrm{5}} \:+\:\mathrm{3}{t}^{\:\mathrm{4}} \:−\:\mathrm{20}{t}^{\mathrm{3}} \:+\:\mathrm{3}{t}^{\mathrm{2}} \:+\:\mathrm{6}{t}\:+\:\mathrm{1}}\:{dt}\:\:\:\: \\ $$$$ \\ $$

Question Number 221587 Answers: 1 Comments: 0

∫_( 2) ^( 3) ((tan^(− 1) (x))/(1 − x^2 )) dx

$$\int_{\:\mathrm{2}} ^{\:\mathrm{3}} \:\frac{\mathrm{tan}^{−\:\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}\:\:−\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$

Question Number 221586 Answers: 1 Comments: 0

∫ ((sin 2x)/(1 + 3x )) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}\:+\:\mathrm{3}{x}\:}\:{dx} \\ $$$$ \\ $$

Question Number 221585 Answers: 6 Comments: 1

solve for x ∈R (x^3 −6)^3 =x+6

$${solve}\:{for}\:{x}\:\in{R} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{6}\right)^{\mathrm{3}} ={x}+\mathrm{6} \\ $$

Question Number 221583 Answers: 1 Comments: 0

Question Number 221578 Answers: 1 Comments: 0

(3+cos x)^2 = 4−2sin^8 x x ε [ 0, 2025π ]

$$\:\: \\ $$$$ \left(\mathrm{3}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} \:=\:\mathrm{4}−\mathrm{2sin}\:^{\mathrm{8}} \mathrm{x}\: \\ $$$$\:\:\mathrm{x}\:\epsilon\:\left[\:\mathrm{0},\:\mathrm{2025}\pi\:\right]\: \\ $$

Question Number 221576 Answers: 0 Comments: 0

Question Number 221577 Answers: 1 Comments: 0

Prove; ∫_0 ^( 1) ((xdx)/((x^2 + 1)(e^(2πx) − 1))) = (γ/2) − (1/4) where; γ is a Euler′s Mascheroni constant

$$ \\ $$$$\:\:\:\:\:\:\mathrm{Prove};\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}\mathrm{d}{x}}{\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({e}^{\mathrm{2}\pi{x}} \:−\:\mathrm{1}\right)}\:=\:\frac{\gamma}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\: \\ $$$$\:\:\:\:\mathrm{where};\:\gamma\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{Mascheroni}\:\mathrm{constant}\:\:\:\: \\ $$$$ \\ $$

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