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... Chris =

**C**So basically: J +**K**= 19**K**+**C**= 14 J +**C**=**K**1) Use substitution postulate: it means...1 Answers · Science & Mathematics · 09/02/2009

When ever I am confronted with a problem like this I refer to my hat size," 6-7/8".

2 Answers · Science & Mathematics · 17/12/2007

...when there are no zeros b^2 - 4ac < 0 b = -4 a =

**k****c**=**k**(-4)^2 - 4(**k**)(**k**) < 0**k**^2 > -4 For what number...1 Answers · Science & Mathematics · 26/02/2014

...solution is y(x) =

**c***e^(6i*x) +**k***e^(-6i*x) +**C**. Now y(0) =**c**+**k**+**C**= -6. Next, y'(x) = 6i***c***e^(6i*x) - 6i***k***e^(-6i*x), so y...1 Answers · Science & Mathematics · 24/02/2013

add both equation to eliminate y (5+

**k**) = 0**k**= -54 Answers · Science & Mathematics · 12/07/2009

Let

**C**(n,**k**) be defined as n!/[**k**! (n-**k**)!] 1.**C**(8,1) +**C**(8,2) + ... +**C**...1 Answers · Science & Mathematics · 22/09/2007

**C**(t) = C0 exp(**k**t)**C**(3.8) = C0/2 = C0 exp(3.8**k**) => 1/2 = exp(3.8**k**) ln(1/2...1 Answers · Science & Mathematics · 27/10/2012

...,

**k**/B) =**C**/A***k**/A -**C**/B***k**/B =**C*****k**-**C*****k**= 0 So now we only have to show that if we start at point (x1...1 Answers · Science & Mathematics · 19/09/2009

Remember that x is a function of t; so

**k**and**c**are constants. So, x(t) = 1/(**k**(t -**c**)) = (1/**k**)(t -**c**)^(-1...2 Answers · Science & Mathematics · 17/10/2012

dx ⁄ dt = [ax² ⁄ (2b)] +

**c**...**k**= a ⁄ (2b) dx ⁄ dt = kx² +**c**... [1 ⁄ (**k**• A)] • arctan[x ⁄ A] + G...1 Answers · Science & Mathematics · 22/10/2010

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