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**k**= 0 is clearly a value for which the system has infinitely many solutions. Suppose**k**≠ 0. --**k**-**k**0**k**0 1 1-**k**0**k**-4**k**-1 3 -- 1 -1 0.... We still get no solutions if**k**= 1. Hence, we get infinite solutions if**k**...1 Answers · Education & Reference · 02/07/2013

**k**=1 , 3/2**k**=2 , 5/8**k**=3 , 10/26 the serie convergens to the asymptote 2^**k**/ 3^**k**= (2/3)^**k**and for**k**-->oo , (2/3)^**k**=0 couse for every n , |n|<1 it will be convergens to 02 Answers · Science & Mathematics · 28/03/2011

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**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

Since

**k**is a natural number, then it's either even or odd. If**k**...then we can let**k**= 2m for some other natural number "m". Then**k**(**k**+1) = 2 * m(**k**+1), so 2 clearly divides it. If**k**is odd...1 Answers · Science & Mathematics · 21/02/2011

**k**= 9 Note that: z^2 - 6x + 9 = (z - 3)^2, so it is a perfect square To find**k**, just divide b = -6 by 2, and then square it. Hope that helps.2 Answers · Science & Mathematics · 19/12/2008

**k**^x = e^xln(**k**) and has no Fourier transform**k**^-x = e^-xln(**k**) has the transform (√(2/π))[(ln(**k**))/(((ln(**k**))^2 + ω^2)] ,**k**> 1, x > 02 Answers · Science & Mathematics · 15/08/2007

**k*** (220)^x = 0.044 or (220)^x = 0.044 /**k**convert to logarithmic formula x = log_220( 0.044 /**k**) convert to base 10 log x = log_10( 0.044 /**k**) / log_10(220) or x = log_10( 0.044 /**k**) / 2.34 hence x = (1/2.34)log_10( 0.044 /**k**)3 Answers · Science & Mathematics · 02/05/2013

C(p,

**k**) = p! /**k**! (p-**k**)!. So C(p,**k**)**k**! (p-**k**)! = p!. Consider the unique factorization in prime numbers of C(p,**k**)**k**! (p-**k**)! = p!. The prime factors of C(p,**k**)**k**! (p-**k**...4 Answers · Science & Mathematics · 18/09/2007

**K**^2[1 + (dy/dx)^2] = (dy/dx)^2 (dy/dx)^2 (**K**^2 - 1) = -**K**^2 dy/dx =**K**/√(1-**K**^2) thus y = Kx/√(1-**K**^2) + C2 Answers · Science & Mathematics · 04/09/2009

**k**^2 + 9**k**+ 14 > 0 (**k**+ 2)(**k**+ 7) > 0**k**> -2 or**k**< -7 in the interval notation : (-∞, -7) U (-2, ∞)2 Answers · Education & Reference · 11/10/2013

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