(0, -3) is not in any quadrant. I lies on the border between two quadrants. Because zero is not positive or negative, it cannot be defined as in a quadrant
(3 , -7) is in the fourth (IV) quadrant.
Origin
The value of x will be negative in the bottom left quadrant (quadrant 3) and the top left quadrant (quadrant 2).
Y = 2/3X - 3 you know Y, 0 out Y to find X intercept 2/3X - 3 = 0 2/3X = 3 X = 9/2 a line between Y = -3 and X = 9/2 comes from the 3rd quadrant, through the 4th quadrant and into the first
there is quadrant 1 , quadrant 2 , quadrant 3 , and quadrant 4
Points on the x-axis or y-axis are not in any quadrant. Therefore, (-3,0) is not contained in a quadrant.
there is no quadrant 0... just 1 2 3 and 4.
(0-5) lies in First Quadrant
what quadrant is (0,-15) in
0
13
Yes, they could. If x+a < 0 and y+b <0.
Quadrant 1: (1,5) Quadrant 2: (-2,3) Quadrant 3: (-3,-3) Quadrant 4:(4,-1)
-3
The coordinate or Cartesian plane is divided into four quadrants by the axes. The axes, themselves, do not belong to any quadrant. Assuming the normal x and y-axes, Quadrant I : x > 0, y > 0 Quadrant II : X < 0, y > 0 Quadrant III : X < 0, y < 0 Quadrant IV: X > 0, y < 0 That's it. No special sides, nothing to solve.
0.75
I can not graph for you, but two points can be found. Zero out X and Y. -X - 3 = 0 -X = 3 X = - 3 Y = - 3, of course Draw a line from the second quadrant into the third quadrant and through the fourth quadrant connecting these two points into a descending line.