Assuming that the numbers are right next to each other, we have
(x)*(x+1)=156, expanding into x^2+x=156 and then x^2+x-156=0. We can then use the quadratic formula to get x=(-1+-sqrt(1+624))/2=(-1+-25)/2= 12, -13. -13 wouldn't be a page number, so 12 must be one number, and 12+1 (13) is the other!
A Nonogon has a whopping 9 sides
hope this helps! :D
Thought you'd want to know: If you're talking about parabolas, it's parabolas, not probables. ;)
The standard equation of a a quadratic is y = ax^2 + bx + c. We need to find the values of the coefficients a, b and c.
Taking the first point: When x=3, y=0, so write 0 = a(3)^2 + b(3) + c, or
0 = 9a + 3b + 1c
Do the same for points (-2,3) and (-1,4).
You will have obtained three linear equations in a, b and c:
3= a(-2)^2 + b(-2) + c, or 3 = 4a - 2b + 1c, also
4 = a(-1)^2 + b(-1) + 1c, or 1a - 1b + 1c.
I used matrix operations to solve this system. The results are:
a= -2/5, b= 1/5, c= 21/5
and so the function f(x) is f(x) = (-2/5)x^2 + (1/5)x + 21/5.
