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Question Number 200446 Answers: 1 Comments: 0

fund Σ_(n=1) ^∞ (((3 + (−1)^n )^n )/n) x^n = ?

$${fund}\:\:\:\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{3}\:+\:\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \right)^{\boldsymbol{{n}}} }{{n}}\:{x}^{\boldsymbol{{n}}} \:=\:? \\ $$

Question Number 200444 Answers: 1 Comments: 0

Question Number 200438 Answers: 0 Comments: 1

Find: (√(3 + (√(6 + (√(9 + ... + (√(96 + (√(99)))))))))) = ?

$$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{3}\:+\:\sqrt{\mathrm{6}\:+\:\sqrt{\mathrm{9}\:+\:...\:+\:\sqrt{\mathrm{96}\:+\:\sqrt{\mathrm{99}}}}}}\:=\:? \\ $$

Question Number 200432 Answers: 0 Comments: 2

Question Number 200428 Answers: 1 Comments: 0

Question Number 200424 Answers: 0 Comments: 4

Question Number 200423 Answers: 0 Comments: 0

Question Number 200420 Answers: 0 Comments: 6

Question Number 200417 Answers: 1 Comments: 0

find: Ω = ∫_1 ^( ∞) ((√x)/((1 + x^2 ))) dx = ?

$$\mathrm{find}:\:\:\:\Omega\:=\:\int_{\mathrm{1}} ^{\:\infty} \:\frac{\sqrt{\mathrm{x}}}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)}\:\mathrm{dx}\:=\:? \\ $$

Question Number 200415 Answers: 1 Comments: 0

Question Number 200413 Answers: 1 Comments: 1

Question Number 200412 Answers: 1 Comments: 0

If (√((x-6)^2 + (y+1)^2 )) + (√((x-9)^2 + (y+5)^2 )) find: minumum = ?

$$\mathrm{If} \\ $$$$\sqrt{\left(\mathrm{x}-\mathrm{6}\right)^{\mathrm{2}} \:+\:\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} }\:+\:\sqrt{\left(\mathrm{x}-\mathrm{9}\right)^{\mathrm{2}} \:+\:\left(\mathrm{y}+\mathrm{5}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{find}:\:\:\:\mathrm{minumum}\:=\:? \\ $$

Question Number 200395 Answers: 1 Comments: 0

Question Number 200388 Answers: 1 Comments: 0

find all values for k such that the eq. x^3 −13x+k=0 has three integer roots.

$${find}\:{all}\:{values}\:{for}\:{k}\:{such}\:{that}\:{the}\:{eq}. \\ $$$${x}^{\mathrm{3}} −\mathrm{13}{x}+{k}=\mathrm{0}\:{has}\:{three}\:{integer}\:{roots}. \\ $$

Question Number 200403 Answers: 1 Comments: 0

∫ (((x^2 + 1)dx)/(x(x−1)(x+1))) = ??

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\left(\boldsymbol{{x}}^{\mathrm{2}} \:+\:\:\mathrm{1}\right)\boldsymbol{{dx}}}{\boldsymbol{{x}}\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}+\mathrm{1}\right)}\:=\:?? \\ $$$$ \\ $$

Question Number 200418 Answers: 1 Comments: 0

calculus ( I ) If , I = ∫_0 ^( π) (( x )/(1 + sin^2 (x))) dx = a ζ ( 2 ) ⇒ a = ? where , ζ (s ) = Σ_(n=1) ^∞ (( 1)/n^( s) )

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{calculus}\:\:\left(\:\:\mathrm{I}\:\:\right)\:\: \\ $$$$\:\:\mathrm{I}{f}\:,\:\:\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\:\pi} \:\frac{\:{x}\:}{\mathrm{1}\:\:+\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)}\:\mathrm{d}{x}\:=\:{a}\:\zeta\:\left(\:\mathrm{2}\:\right)\:\: \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\:\:{a}\:=\:?\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:{where}\:\:,\:\:\:\zeta\:\left({s}\:\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\mathrm{1}}{{n}^{\:{s}} } \\ $$$$ \\ $$

Question Number 200379 Answers: 0 Comments: 1

Question Number 200377 Answers: 1 Comments: 1

Question Number 200366 Answers: 1 Comments: 1

Question Number 200357 Answers: 0 Comments: 4

Question Number 200356 Answers: 1 Comments: 1

Question Number 200353 Answers: 3 Comments: 0

If the roots of x^3 +3px^2 +3qx+r=0 are in harmonic progression, then prove that 2q^3 =r(3pq−r)

$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} +\mathrm{3px}^{\mathrm{2}} +\mathrm{3qx}+\mathrm{r}=\mathrm{0}\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{harmonic}\:\mathrm{progression},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{2q}^{\mathrm{3}} =\mathrm{r}\left(\mathrm{3pq}−\mathrm{r}\right) \\ $$

Question Number 200351 Answers: 1 Comments: 0

Solve the equation x^3 −12x^2 −6x−10=0 by cardon′s method

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{3}} −\mathrm{12x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{10}=\mathrm{0}\: \\ $$$$\mathrm{by}\:\mathrm{cardon}'\mathrm{s}\:\mathrm{method} \\ $$

Question Number 200350 Answers: 0 Comments: 1

Question Number 200336 Answers: 1 Comments: 0

Question Number 200330 Answers: 4 Comments: 0

2^x − 3^x = (√(6^x − 9^x )) find: x = ?

$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:=\:\sqrt{\mathrm{6}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{9}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}\:=\:? \\ $$

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