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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

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

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

**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

**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

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

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**| = 33 Answers · Science & Mathematics · 25/03/2014

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

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

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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