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March 8, 2014

# Search: alberbra

Number of results: 8

alberbra
3, thanx
Sunday, July 27, 2008 at 7:57am by Alex

alberbra
thank-you very much
Sunday, July 27, 2008 at 7:57am by Alex

Alberbra
Please help with graphing a quadratic function. y=(x-1)^2 Got stuck at y=(x-1)(x-1) and p=1, q=1 Which doesn't work, because you can't fix a vertex with one point on a graph.
Friday, January 1, 2010 at 1:19pm by Kelly

alberbra
first of all, find the number of days in 4 years, which would be 3x365 + 366 (one leap year) the days that a law can be passed has to be divisible by 10, 12 and 7 the lowest common multiple would be 420 How many multiples of 420 are there in your total number of days?
Sunday, July 27, 2008 at 7:57am by Reiny

Algebra
What is Alberbra? I don't know what your p and q are upposed to be. Define a new variable x' = x-1. Your equation becomes y = x'^2 The vertex of your parabola is at x=1, y=0. That is one point of the graph. Other points are easily plotted. The parameter p is often used to ...
Friday, January 1, 2010 at 1:19pm by drwls

alberbra
The number of days after January 1, 1996 must be divisible by 10, 12 and 7. It must also be less than 365x4 = 1460. The Lowest Common Multiple of those numbers is 2^2*3*5*7 = 420 (i.e, they can pass a law every 60 weeks.)The number of opportunitites is the integer part of 1460...
Sunday, July 27, 2008 at 7:57am by drwls

Alberbra
The standard quadratic is y=a(x-h)²+k where (h,k) is the vertex, and x=h is the axis of symmetry. The zeroes (p,q) are at -h±√(k/a). If a>0, the quadratic is concave upwards, i.e. the vertex is at the minimum. If a<0, the graph is concave downwards, or ...
Friday, January 1, 2010 at 1:19pm by MathMate

alberbra
The parliament of the land of Archronia consists of two houses. The Parliament was elected in 1995 for a period of four years beginning on Monday, January 1st, 1996, when the two houses had their first sessions. According to the rules, the meetings of the first house must ...
Sunday, July 27, 2008 at 7:57am by Alex

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