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**k**...**k**= (ln|y - 90| -**c**)/t**c**....**c**= ln|y - 90| - kt I don't get the initial values since you have more than two constants,**k**and**c**.2 Answers · Education & Reference · 12/09/2011

... think Katharine Krampf sounds lovely!! I love

**K**'s and**C**'s too and this name flows very well with your last name! :D...5 Answers · Pregnancy & Parenting · 09/11/2012

Ok. 273

**K**= 0**C**so therefore 330C = 330 + 273 = 603K**C**.5 Answers · Science & Mathematics · 21/06/2010

By the Binomial Theorem, (1 + x)^n = sum(

**k**=0 to n)**C**(n,**k**) x^**k**. Differentiating both sides twice, we get n(n - 1) (1 + x...1 Answers · Science & Mathematics · 19/10/2009

...28 +

**c c**= 6k - 10 12k = 28 + 6k - 10 6k = 18**k**= 3**c**= 6(3) - 10**c**= 8**k**+**c**= 3 + 8 = 111 Answers · Science & Mathematics · 11/01/2009

n *

**C**(n-1,**k**-1) = n * [(n-1)! / ((**k**-1)! (n-**k**)!)] = n! / ((**k**-1)! (n-**k**)!) =**k*** [n! / (**k**! (n-**k**)!)] =**k*****C**(n,**k**). I hope this helps!1 Answers · Science & Mathematics · 20/11/2011

ƒ( k ) = √[ k + ( k − c )^2], assuming c is a constant, and k is the only variable. ƒ...v, by chain rule, we get: ƒ'( k ) = u' ⋅ v ⋅ u^(v - 1) ƒ'( k ) = [1 + 2( k − c )] ⋅ 1/2 ⋅ [ k + ( k − c )^2]^(-1/2) ƒ'( k ) = (1 + 2k − 2c) / (2√[ k + ( k − c ...

3 Answers · Science & Mathematics · 23/02/2013

A = 4k = 0i + 0j + 4k B = 8i + k = 8i + 0j + k C = 3i - 4j = 3i - 4j + 0k Now, (A • B) = (0*8 + 0*0 + 4*1) = 4...

2 Answers · Science & Mathematics · 24/09/2012

...respectively. Reference (2) gives formulas for h=-b/2a and

**k**=**c**- [(b^2)[(4a)]. Using these we answer the questions that h = -(-12)/[2...1 Answers · Science & Mathematics · 21/02/2013

..., but that'd just be ridiculous). However, the answer is: ° C = ° K - 273.16 ° C = 300 - 273,16 = 26.84° C , or, rounded off, to 27...

3 Answers · Science & Mathematics · 08/02/2008

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