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f(2) = 2² + k (2) + c = 0 f(-3) = -3² + k (-3) + c = 35 Drop the functions on the left and multiply...4 - 12 + 8 = 0 9 + 18 + 8 = 35 -8 + 8 = 0 27 + 8 = 35 0 = 0 35 = 35 This checks. k = -6 and c = 8

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... in the formula: since

**k**= n - (n-**k**), we have:**C**(n,**k**) = n! / (**k**! (n-**k**)! )**C**(n, n-**k**) = n! / ((n-**k**)! (n - (n-**k**))! = n! / ( (n-**k**)!**k**! )1 Answers · Science & Mathematics · 14/03/2012

See binomial equation [1 + x]^n = C (n,0)+ x* C (n,1) +(x^2)* C (n,2)+ (x^3)* C (n,3) ... + [x^n-1...(n,1)+ (2*x)* C (n,2) + (3x^2)* C (n,3)+ ... + (n-1)*[x^n-2]⋅ C (n,n-1) + n*[x^n-1]* C (n,n) put x = 1, you will see that right part is...

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Assuming that

**k**and**c**are just constants, the equation is separable, divide both ...subject, divide both sides by -1 and raise both sides to the base of e:**c**- y = e^(-kx) * e^-**C**As e^-**C**represents a constant...3 Answers · Science & Mathematics · 17/08/2013

...form one mole gaseeous atoms. now considering Na,

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