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1. ### Math question!!!!!!!!!!!!!!!?

... 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

2. ### log base b (a)/ log base c (a)= 19/99 and b/c =c^k find k.?

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

3. ### For what value(s) of k will the function f(x)=kx^2 -4x + k have no zeros?

...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

4. ### find y as a function of x if y'''+36y'=0 y(0)=-6 , y'(0)=-18, y''(0)=-36?

...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

5. ### Find a value for k such that the following system will have exactly one solution.?

add both equation to eliminate y (5+k) = 0 k = -5

4 Answers · Science & Mathematics · 12/07/2009

6. ### Data Management Question?

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

7. ### CALCULUS!!!!!!!!!!!!!!!!!!!!!!!!!!!?

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

8. ### Show that the distance between the point (x1,y1) and the line Ax+Bx+C=0 is?

...,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

9. ### Show that the function satisfies the differential equation.....?

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

10. ### (dx/dt) = (ax^2)/2b + c?

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