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  1. If k = 1, then this series diverges by the nth term test, because lim(n→∞) n! = ∞. ---------- Assuming k > 1, now we use the Ratio Test: r = lim(n→∞) [((n+1)!)^2 / ( k ...

    1 Answers · Science & Mathematics · 07/04/2013

  2. y = 2x + k ← this is a line y = x² + kx + 5 ← this is a curve y = y → 2 points → 2 solutions for x x² + kx + 5 = 2x + k x² + kx + 5 - 2x - k = 0 x² + x.( k - 2) + (5 - k ) = 0 Polynomial...

    3 Answers · Science & Mathematics · 22/01/2016

  3. f(x) = k - x² ← this is the curve y = 4x - 9 ← this is the line ... us only one point, so only one solution to the equation 4x - 9 = k - x² x² + 4x - 9 - k = 0 x² + 4x - (9 + k ) = 0 ...

    4 Answers · Science & Mathematics · 24/09/2015

  4. a[ k ] = 1/ln( k )^ln( k ) a[ k +1] = 1/ln( k (1+1/ k ))^ln( k +1) = 1/( ln( k ) + ln(1+1/ k ))^ln( k +1) = 1/ ( ln( k )^ln( k +1) ( 1+ (ln(1+1/ k )/ln( k ) )^ln( k +1) a[ k ]/a[ k +1] =ln( k )(1+(ln...

    3 Answers · Science & Mathematics · 18/04/2007

  5. "Solve for k in terms of k " doesn't make sense: k = k is obvious. I...thing and see if we can get a cleaner statement of the relationship between x and k : log(base2)x+log(base2)(x+5)= k log(base2) [x(x+5)] = k {by properties of...

    2 Answers · Science & Mathematics · 19/10/2009

  6. ========= x^2 + k + 6 = kx +3x x^2 -x( k +3) + ( k +6) = 0 The equation will have equal ... happens when the discriminant is zero. So, solve ( k +3)^2 - 4( k +6) = 0 k ^2 + 6k + 9 - 4k - 24 = 0 k ^2 + 2k - 15 = 0 ( k +5)( k -3...

    2 Answers · Science & Mathematics · 14/10/2012

  7. This only true if K is also a metric space. Otherwise there are counterexamples.... Now to prove the claim: Forward direction: Suppose K is compact and A = (x_n) is a sequence in K . We...

    2 Answers · Science & Mathematics · 16/10/2010

  8. k =ideal gas constant k =universal gas constant k =constant proportional k =0.08205 L.atm/mol. K (if you're gonna use liter) or k =82.05ml.atm/mol. K (if you're gonna use ml) k is also be R it ...

    1 Answers · Science & Mathematics · 28/04/2008

  9. ... 6/( k +1) - 1/( k ) = 1 ← note: k ≠ 0 or -1 or 6( k ) - ( k +1) = ( k +1)( k ) or 6k - k - 1 = k² + k or 0 = k² + k - 6k + k ...4*3 ) ] / 2 k = [ 4 ± 2√3 ) ] / 2 k = 2 ± √3 k = { 2 - √3, 2 + √3 } k ~ { 0.2679, 3.7321 }

    3 Answers · Science & Mathematics · 11/09/2010

  10. k = ( k +15)/3*( k -1) ==> multiply both sides by 3( k -1) to get rid of fractions 3k( k -1) = k +15 ==> expand the LHS 3k^2-3k = k +15 ==> move everything to the LHS 3k^2-4k-15=0 actually this does...

    3 Answers · Science & Mathematics · 21/07/2007

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