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There is no equation in the question, only the expression y6x.
II and IV
A curved line can pass through (not threw) all four quadrants. The maximim for a straight line is three.
I would say from an educated guess that it is 0. A straight line could avoid all quadrants if it were placed on the origins of the x and y axis.
They satisfy the equation x + y = 0
At what point does line represented by the equation 8x + 4y = -4 intersects the y-axis, and at what point in the negative direction of x-axis.
I,ii
II and IV
Quadrants I and III, numbered from I at upper right (+, +) left and moving clockwise. The line passes through the origin (0,0).
It will pass through the first (when x is positive) and third quadrants (when x is negative, y will also be negative).
A curved line can pass through (not threw) all four quadrants. The maximim for a straight line is three.
I would say from an educated guess that it is 0. A straight line could avoid all quadrants if it were placed on the origins of the x and y axis.
They satisfy the equation x + y = 0
At what point does line represented by the equation 8x + 4y = -4 intersects the y-axis, and at what point in the negative direction of x-axis.
What is the equation of the vertical line passing through (-5,-2)
It intercepts the y axis at (0, 5) and it intercepts the x axis at (-2.3, 0) passing through the I, II and III quadrants
The equation of a horizontal line is of the form y=k, where k is the y-coordinate of the point through which the line passes. Therefore, the equation of the horizontal line through the point (8, -10) is y = -10.
y = 0. You can get this from the slope-intercept equation of the line.