y=x^3
The slope of the line that represents the graph of that equation is 15.
A linear graph is a model of a straight line on the X and Y axis. It represents the equation y=mx+b. A liner graph has a slope. A liner graph cannot be equaled to 0.
The is a straight line parallel to the y-axis with an x intercept at -3.
If there is one equation that can be solved, then (x-2)2 + (y-3)2 = 0 For an equation such as x+y-5 = 0 you will have an infinity of solutions unless you add another, independent equation and solve them simultaneously.
The graph of is shifted 3 units down and 2 units right. Which equation represents the new graph?
To graph the equation y-x=3, first rearrange it in slope-intercept form by isolating y: y=x+3. This equation represents a line with a slope of 1 and y-intercept of 3. You can plot the y-intercept at (0,3) then use the slope to find another point and draw a straight line connecting the two points.
Move 3 over the right side of the equation so the equation would be x = -3. The graph of this would be a verticle line at x= -3
yx-3 is not an equation, and it has no graph.
Assuming the graph is: x = - 305 y, then the y intercept is at y = 0Assuming the graph is: x - 3 = 5y, then the y intercept is at y = -3/5The main point for you to realize here is that a graph represents an equation,and " -305y " is not an equation. So some kind of assumption has to be madein order to come up with something that can be graphed.
y=x^3
The slope of the line that represents the graph of that equation is 15.
A linear graph is a model of a straight line on the X and Y axis. It represents the equation y=mx+b. A liner graph has a slope. A liner graph cannot be equaled to 0.
The is a straight line parallel to the y-axis with an x intercept at -3.
If there is one equation that can be solved, then (x-2)2 + (y-3)2 = 0 For an equation such as x+y-5 = 0 you will have an infinity of solutions unless you add another, independent equation and solve them simultaneously.
False. X = 3 is a vertical line.
Equation of a circle when its centre is at (0, 0): x^2 + y^2 = radius^2 Equation of a circle when its centre is at (a, b): (x-a)^2 + (y-b)^2 = radius^2