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1. Rate Contant (k)..........?

k = rate/(mol/L)^3 If the rate is in mole/sec, then k = L^3/(mole.sec)

4 Answers · Science & Mathematics · 17/05/2011

2. what is 5k divided by -k?

k-8/7+k = -1/5 okay :P afraid? maths <3 awesome :P love XD 5(k-8) = -1 (7+k) 5k - 40 = -7 - k 5k+k= -7 + 40 6k = 33 k= 33/6

2 Answers · Education & Reference · 07/10/2012

3. Show that p_k = p_(k-1) + (3k-2) for k >=2. Conclude that p_n = sum from k=1 to n of (3k-2).?

Let k ≥ 2. Then p_(k - 1) + 3k - 2 = (k - 1)[3(k - 1) - 1]/2 + (6k - 4)/2 = [(k - 1)(3k - 3 - 1) + 6k - 4]/2 = [(k - 1)(3k - 4) + 6k - 4]/2 = (3k² - 7k...

1 Answers · Science & Mathematics · 13/09/2009

4. Find the values of k?

...1 ... equation 1 (given)     - x  +         y  +  kz  =  k ... equation 3 (given)    ——————————<ADD...

2 Answers · Education & Reference · 24/04/2013

5. k-4 divided by 9=3 help me?

k-4/9=3 you want to isolate k to find the answer, so you have to do the operations in reverse. so k-4=3x9 k-4=27 k=27+4 k=31

3 Answers · Education & Reference · 28/02/2010

6. For what values of the constant k will the following function NOT be its own inverse.?

Let (x - k) / (x - 1) = y x - k = yx - y x = yx - y + k x - yx = y + k x(1 - y) = y + k x = (-y + k) / (1 - y) x = (y - k) / (y - 1) It would seem that the function is generally its own inverse...

2 Answers · Science & Mathematics · 12/08/2010

7. What is so special about Special K??

Depends which Special K you mean....... if your enquiry is in regards to cereal then the...

14 Answers · Entertainment & Music · 11/01/2008

8. please help me solve y = k ^ .3?

k^.3 is the same as saying k^(3/10) k^(3/10) is the same as saying tenth_root(k^3) k...

7 Answers · Science & Mathematics · 04/01/2007

9. From the formula, k(x)=2sin(x+pi)/(sin(x+pi)-3), what is k^-1(x)?

k(x) = 2 sin (x+π) / (sin (x+π) - 3) = 2 (- sin x) / (- sin x - 3) = 2 sin x / (sin x + 3) = [2 (sin x + 3) - 6] / (sin x + 3) = 2 - 6 / (sin x + 3) So for k^-1(x) we get x = 2 - 6 / (sin k^-1(x) + 3) => 6 / (sin k^-1(x) + 3) = 2 - x =>...

1 Answers · Science & Mathematics · 16/12/2013

10. Solving Polynomial Equations k(k-5)>0?

k(k - 5) > 0 in two cases: 1) k > 0 and (k - 5) > 0 2) k < 0 and (k - 5) < 0 In case 1, k > 0 and k > 5; both of these are true when k > 5. In case 2, k...

1 Answers · Science & Mathematics · 11/04/2010