x,y = 5,6 or x = 5 and y = 6
You need two, or more, curves for points of intersection.
It works out that the points of intersection between the equations of 2x+5 = 5 and x^2 -y^2 = 3 are at: (14/3, -13/3) and (2, 1)
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
x,y = 5,6 or x = 5 and y = 6
You need two, or more, curves for points of intersection.
It works out that the points of intersection between the equations of 2x+5 = 5 and x^2 -y^2 = 3 are at: (14/3, -13/3) and (2, 1)
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
+ (a plus sign) is the symbol for an intersection.
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
2 points plus
FOUR
how many weight watchers points plus are in a meatball in sauce
how many points for mussels
I weigh 141 pounds and my points are 29. 22 points for WW Plus is NOT correct.
I would say zero points.