y - 2 = 5 is the same as y = 7, which is a horizontal line above the x-axis at y=7.
Straight lines.
y-intercept is -2, so start at the point (0,-2) slope is 1 so move up 1 and over 1. use a straight edge to connect the dots and continue the line.
1
(y^2 - 4) / (y + 2) = [(y -2)(y+2)]/(y+2) = y-2
It isn't supposed to have dots. Usually dots on coins are made by nicks in the die.
above are examples. You need a y and x axis (labelled), data (dots) and the data joined up by a line. Like a scatter graph but with the dots joined up
y - 2 = 5 is the same as y = 7, which is a horizontal line above the x-axis at y=7.
y + x = -2 x = -y - 2Plug in numbers for y and solve for x. Then graph the x's and y's, and connect the dots. The x direction is right tot left. The y direction is up and down. So if y = -1, then x = -1. If y = 1 then x=-3.
Y = 2 The graph is a horizontal line passing through the point Y=2 on the Y=axis. The line is parallel to the X-axis, and exactly 2 units above it everywhere.
Straight lines.
y-intercept is -2, so start at the point (0,-2) slope is 1 so move up 1 and over 1. use a straight edge to connect the dots and continue the line.
Any two numbers of this type can be solved using boolean logics....... It is very simple once had a glance on it . Here is a method to prove x=y where x and y are any numbers..... let XY = XY We can rewrite the above as ð X2 - (X+Y)*X = Y2 - (X+Y)*Y ð Multiplying both sides with 2/2 we get ð X2 - 2*x*(X+Y)/2 = Y2 - 2*Y*(X+Y)/2 ð Adding (X+Y)2/2 on both sides we get ð X2 - 2*x*(X+Y)/2 + (X+Y)2/2= Y2 - 2*Y*(X+Y)/2 + (X+Y)2/2 ð This is in the form of a2-2ab +b2 = (a+b)2 ð (X - (X+Y)/2)2 = (Y-(X+Y)/2)2 ð Canceling the squares on both sides(case condition) ð X-(X+Y)/2 = Y - (X+Y)/2 ð Canceling (X+Y)/2 on both sides ð X=Y
Draw a blank graph (x-y axis), then start plotting points until you find the straight-line (e.g. plot at least 2 points, then get a straight-edge and connect the dots):(x,y)(0,0)(1,-2)(2,-4)(-1,2)(-2,4)
(x - y)2 - z2 is a difference of two squares (DOTS), those of (x-y) and z. So the factorisation is [(x - y) + z]*[(x - y) - z] = (x - y + z)*(x - y - z)
Positive x- and y-coordinates of a point in the first quadrant.
neither