- k ^3 + k ^2 + 2* k + 4 = - k ( k ^2 - k - 2) + 4 = - k ( k ^2 -2k + k -2) + 4 = - k [ k ( k -2) + 1( k -2)] + 4 = - k ( k +1)( k -2) + 4 How do you want us to calculate? What...
1 Answers · Education & Reference · 25/01/2011
- k ===-1k .+-7k===-8k Take the numbers to the right side by changing their signs: +6-20-2===-16 -8k===-16 ... ................... k ===2 Here is a more step by step explanation: There is...
2 Answers · Education & Reference · 08/09/2013
y = k √x y' = k / 2√x Now if the given line is tangent to y = k √x then the slope of the line equals k / 2√x so k / 2√x = 5 k = 10...
1 Answers · Education & Reference · 16/02/2010
k = -3 or -1 For the general quadratic ax^2 + bx +c = f(x), "solving" means finding the...'re into Complex Numbers System Now, on to your question x^2 -kx - k +1 has a Discriminant of -7 Comparing this to ax^2 + bx +c =0 a = 1, b=- k , c = - k +1 ...
1 Answers · Education & Reference · 03/10/2011
For: aA + bB = cC + dD, K may be defined as; K = (P(C)^c * P(D...
1 Answers · Education & Reference · 24/09/2006
There is no solution. If k = 0 , the equation becomes e^(0) + 4(0) + 1 = 1 + 1 = 2 If k >...; 0 . As a result, the entire expression [ e^(-6k)+4k+1 ] > 0 when k < 0. Here's a view of part of the graph of f( k ) = e...
1 Answers · Education & Reference · 29/04/2012
k shifts the graph vertically. If k is positive the graph shifts up; if k is negative it shifts...
1 Answers · Education & Reference · 02/03/2014
since (1, k ) is a point on the graph of this equation, all you have to do is plug 1 in for x, and when you find y, you've found k . y = 2(1)^2 - 4(1) + 6 = 2 - 4 + 6 = 4 k = 4 that's it! ;)
2 Answers · Education & Reference · 14/05/2012
Suppose k = 1. Then you have x^(2^1) = x^2, which involves one act of ...2) = x^4, which involves two acts of squaring. If k = 3, then you have x^(2^3) = x^8, which takes three. See the pattern...
3 Answers · Education & Reference · 24/10/2007
|---- k ----|---- 2k ----|---- 3k ----|---- 4k ----|--> With the timeline as above, your solution seems correct. 10 000 = k a[5] + 2k a[5] * v^5 + 3k a[5] * v^10 + 4k a[5] * v^15 = k a[5] * (1 + 2 v^5...
1 Answers · Education & Reference · 06/11/2011