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Question Number 90135    Answers: 1   Comments: 0

I_n =∫_(t=0) ^(+∞) (dt/((t+1)(t+2)...(t+n)))

$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{t}=\mathrm{0}} ^{+\infty} \frac{\mathrm{dt}}{\left(\mathrm{t}+\mathrm{1}\right)\left(\mathrm{t}+\mathrm{2}\right)...\left(\mathrm{t}+\mathrm{n}\right)} \\ $$

Question Number 90134    Answers: 0   Comments: 0

Question Number 90122    Answers: 0   Comments: 1

each vertex of a cube is to be labeled with an integer 1 through 8 , with each integer being used once,in such a way that the sum of the four numbers on the vertices of a face is the same for each face.Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. Find all the arrangements

$${each}\:{vertex}\:{of}\:{a}\:{cube}\:{is}\:{to}\:{be}\:{labeled} \\ $$$${with}\:{an}\:{integer}\:\mathrm{1}\:{through}\:\mathrm{8}\:,\:{with} \\ $$$${each}\:{integer}\:{being}\:{used}\:{once},{in}\:{such} \\ $$$${a}\:{way}\:{that}\:{the}\:{sum}\:{of}\:{the}\:{four}\:{numbers} \\ $$$${on}\:{the}\:{vertices}\:{of}\:{a}\:{face}\:{is}\:{the}\:{same}\: \\ $$$${for}\:{each}\:{face}.{Arrangements}\:{that}\:{can} \\ $$$${be}\:{obtained}\:{from}\:{each}\:{other}\:{through} \\ $$$${rotations}\:{of}\:{the}\:{cube}\:{are}\:{considered} \\ $$$${to}\:{be}\:{the}\:{same}.\: \\ $$$${Find}\:{all}\:{the}\:{arrangements} \\ $$$$ \\ $$$$ \\ $$

Question Number 90162    Answers: 1   Comments: 0

Question Number 90160    Answers: 1   Comments: 0

Question Number 90161    Answers: 1   Comments: 0

Question Number 90113    Answers: 1   Comments: 0

∫_0 ^∞ e^(−x) ((1/(1−e^(−x) ))−(1/x))dx

$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} \left(\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{x}} }−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$

Question Number 90110    Answers: 0   Comments: 2

∫_((2/3)u) ^(2u) (e^(−(x/2)) /(2π (√((u−(1/2)x)(((3x)/2)−u))))) du (u > 0 )

$$\underset{\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}} {\overset{\mathrm{2u}} {\int}}\:\frac{\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{2}}} }{\mathrm{2}\pi\:\sqrt{\left(\mathrm{u}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)\left(\frac{\mathrm{3x}}{\mathrm{2}}−\mathrm{u}\right)}}\:\mathrm{du}\: \\ $$$$\left(\mathrm{u}\:>\:\mathrm{0}\:\right) \\ $$

Question Number 90103    Answers: 0   Comments: 1

Question Number 90100    Answers: 1   Comments: 0

Question Number 90099    Answers: 0   Comments: 2

given the polar equation r = a^2 sin2θ show the tangents at the poles of this polar equation is. θ = {(π/4),((3π)/4),((5π)/4),((7π)/4)}

$$\:\mathrm{given}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{equation} \\ $$$$\:{r}\:=\:{a}^{\mathrm{2}} \:\mathrm{sin2}\theta\:\:\mathrm{show}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at}\: \\ $$$$\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\mathrm{this}\:\mathrm{polar}\:\mathrm{equation}\:\mathrm{is}. \\ $$$$\:\theta\:=\:\left\{\frac{\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}},\frac{\mathrm{7}\pi}{\mathrm{4}}\right\} \\ $$

Question Number 90097    Answers: 0   Comments: 1

((√(3+(√8))))^x +((√(3−(√8))))^x = 6

$$\left(\sqrt{\mathrm{3}+\sqrt{\mathrm{8}}}\right)^{\mathrm{x}} \:+\left(\sqrt{\mathrm{3}−\sqrt{\mathrm{8}}}\right)^{\mathrm{x}} \:=\:\mathrm{6} \\ $$

Question Number 90090    Answers: 1   Comments: 0

xy (dy/dx) = y^2 + ((x^3 /(x^2 +1)))

$${xy}\:\frac{{dy}}{{dx}}\:=\:{y}^{\mathrm{2}} \:+\:\left(\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$

Question Number 90083    Answers: 0   Comments: 0

Question Number 90086    Answers: 0   Comments: 7

Question Number 90087    Answers: 0   Comments: 1

Σ_(k = 1) ^∞ (1/k^k ) = ?

$$\underset{{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{{k}} }\:=\:? \\ $$

Question Number 90080    Answers: 0   Comments: 0

Question Number 90077    Answers: 0   Comments: 3

lim_(x→∞) (sin (x+(1/x))−sin(x))=?

$${lim}_{{x}\rightarrow\infty} \left(\mathrm{sin}\:\left({x}+\frac{\mathrm{1}}{{x}}\right)−{sin}\left({x}\right)\right)=? \\ $$

Question Number 90075    Answers: 0   Comments: 1

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Question Number 90073    Answers: 0   Comments: 0

$$ \\ $$

Question Number 90092    Answers: 0   Comments: 1

G((√(x+5))) = x G(x^2 ) = x^a −b find a+b

$$\mathrm{G}\left(\sqrt{\mathrm{x}+\mathrm{5}}\right)\:=\:\mathrm{x} \\ $$$$\mathrm{G}\left(\mathrm{x}^{\mathrm{2}} \right)\:=\:\mathrm{x}^{\mathrm{a}} −\mathrm{b} \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{b}\: \\ $$

Question Number 90069    Answers: 2   Comments: 3

Question Number 90060    Answers: 0   Comments: 1

lim_(x→0) ((ln (1+sin x))/(((2+x))^(1/(3 )) − ((2+3x))^(1/3) )) = ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)}{\sqrt[{\mathrm{3}\:\:}]{\mathrm{2}+\mathrm{x}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\mathrm{3x}}}\:=\:? \\ $$

Question Number 90058    Answers: 1   Comments: 1

Question Number 90055    Answers: 0   Comments: 1

Question Number 90049    Answers: 0   Comments: 0

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