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  1. Let (k_n) be a sequence in K. Since each k_n is in at least one of... C(x, r), there is a sequence (x_n) in X such that ||k_n - x_n|| ≤ r for all n (1). Since X is compact, (x_n) ...

    3 Answers · Science & Mathematics · 03/12/2013

  2. e^(2x) = k√(x) We first have to assume that, (e^(2x)) = k√(x), has a unique solution. If that's the case then, (e^2x) / √(x) = k, must have a unique solution as well. As a result the unique solution for, (e^2x...

    2 Answers · Science & Mathematics · 24/10/2013

  3. 0=((K)/(s)+1)-((s)/(s)+1) Since K...on the left-hand side of the equation. ((K)/(s)+1)-((s)/(s)+1)=0 To add..., add -2s to s to get -s. (K-s)/(s)=0 Multiply each term...

    2 Answers · Science & Mathematics · 05/01/2013

  4. k^2 < 36 Subtract 36 from each side: k^2 - 36 < 0 Factor: (k + 6) * (k... correct By this, the only valid solution was k = 0, which we took from the (-6, 6) interval. Thus...

    3 Answers · Science & Mathematics · 12/11/2011

  5. k(x)/p(x) is just k(x) divided by p(x) which is (x-2)/(2x^2-5x+2). ... except 2 and 1/2. Now we can simplify to say that k(x)/p(x) = 1/2x-1. It is important to state the domain before...

    1 Answers · Science & Mathematics · 21/06/2013

  6. Note that ∏(k=2 to n) (k³ - 1)/(k³ + 1) = ∏(k=2 to n) [(k - 1)(k² + k...k + 1)(k² - k + 1)] = ∏(k=2 to n) [(k - 1) ((k+1)² - (k+1) + 1)]/[(k + 1)(k² - k + 1)] = {∏(k=2 to n) (k - 1)/(k + 1)} · {∏(k=2 to n) ((k+1...

    1 Answers · Science & Mathematics · 25/10/2010

  7. let k = x+iy then |k| = sqrt(x^2+y^2) given |k+9|=3|k+1| |(x+iy)+9|=3|(x+iy)+1| |(x+9)+iy|=3|(x+1)+iy| sqrt{(x+9)^2+y^2}=3.sqrt{(x...2+18x+9+9y^2 8x^2+8y^2 = 81-9 = 72 x^2+y^2 = 9 sqrt(x^2+y^2) = 3, that is |k| = 3

    3 Answers · Science & Mathematics · 25/03/2014

  8. 1. S(k, a, r) is the sum of the first k terms of the geometric series: a, ar, ar^2, ..., ar^(k-1) 2. tn=100(11/17)^n-1 corresponds to the general term of the geometric...

    1 Answers · Science & Mathematics · 13/04/2009

  9. 1. ∑ k=1. 3(k/n) = (3/n)∑ k=1. (k) = (3/n)(n/2)(n - 1) = (3/2)(n - 1) 2. ∑ k=1. (4+2k)^2 = ∑ k=1. (16 + 16k + 4k^2) = 16n + 16(n/2)(n - 1) + 4[(1/3)n...

    1 Answers · Science & Mathematics · 22/06/2013

  10. x^2 + ((k+1)/k)x + 1 = 0 (x + ((k+1)/(2k)))^2 = -1 + (k+1)^2 / (4k^2) 1) for 1 root, (k+1)^2 / (4k^2) = 1 => (k+1...1 > 0 and k - 1 > 0 or 3k + 1 < 0 and k - 1 < 0 => k > 1 or k <...

    2 Answers · Science & Mathematics · 05/01/2017

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