Assume |

**k**−√2| < ϵ. Need to prove that | (**k**+2)/(**k**+1)−√2... from |**k**−√2| < ϵ assemble an expression for (**k**+2)/(**k**+1)−√2 |**k**−√2| < ϵ → √2−ϵ &...1 Answers · Science & Mathematics · 28/02/2013

... 6/(

**k**+1) - 1/(**k**) = 1 ← note:**k**≠ 0 or -1 or 6(**k**) - (**k**+1) = (**k**+1)(**k**) or 6**k**-**k**- 1 =**k**² +**k**or 0 =**k**² +**k**- 6**k**+**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

"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

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

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

"

**k**multiplied by three": 3k "the sum of**k**multiplied by three and**k**": 3k +**k**"the sum of**k**multiplied by three and**k**is 12": 3k +**k**= 12 3k +**k**= 12 4k = 12**k**= 12/4**k**= 32 Answers · Science & Mathematics · 22/04/2012

It depends on what

**k**can be. Is**k**always a positive integer? Is**k**always a negative integer...a positive integer. Since 1 = cos 0 + i sin 0, we see that, in polar form:**k**=**k**(cos 0 + i sin 0). By DeMovire's Theorem: x...1 Answers · Science & Mathematics · 28/03/2011

========= 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

y = 2x +

**k**← this is a line y = x² +**k**x + 5 ← this is a curve y = y → 2 points → 2 solutions for x x² +**k**x + 5 = 2x +**k**x² +**k**x + 5 - 2x -**k**= 0 x² + x.(**k**- 2) + (5 -**k**) = 0 Polynomial...3 Answers · Science & Mathematics · 22/01/2016

... D between two points = sqrt[(x-x1)^2 +(y-y1)^2] D = sqrt[(

**k**+3)^2 +(**k**-0)^2] = sqrt[(**k**-0)^2 +(**k**-6)^2] sqrt[(**k**+3)^2 +(**k**-0)^2] = sqrt[(**k**-0)^2...4 Answers · Science & Mathematics · 12/09/2010

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