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How do you find the locus of all points that lie on the graphs of both x plus 2y equals 1 and 2x -y equals 12?
You solve the two equations simultaneously. There are several ways to do it; one method is to solve the first equation for "x", then replace that in the second equation. This will give you a value for "y". After solving for "y", replace that in any of the two original equations, and solve the remaining equation for "x".
You solve the two equations simultaneously. There are several ways to do it; one method is to solve the first equation for "x", then replace that in the second equation. This will give you a value for "y". After solving for "y", replace that in any of the two original equations, and solve the remaining equation for "x".
You solve the two equations simultaneously. There are several ways to do it; one method is to solve the first equation for "x", then replace that in the second equation. This will give you a value for "y". After solving for "y", replace that in any of the two original equations, and solve the remaining equation for "x".
You solve the two equations simultaneously. There are several ways to do it; one method is to solve the first equation for "x", then replace that in the second equation. This will give you a value for "y". After solving for "y", replace that in any of the two original equations, and solve the remaining equation for "x".
What else can I help you with?
What is the locus of points equidistant from two parallel lines?
It's a third line, parallel to both and midway between them.
What is the locus of all points in a plane equidistant from two parallel lines in the plane?
It's another line, parallel to both of the first two and midway between them.
How are bar graphs and picture graphs alike?
Both bar graphs and picture graphs show statistics (data) in a visual (graphic) form.
What is the locus of points that satisfies both equations 2x-3y equals 5 and x-2y equals 1?
2x-3y=5 and x-2y=12x-3y+2y=5 and x-2y+2y=12x-y=5 and x=12(1)-y=52-y=52-y-2=5-2-y=3y=-3(x,y) is (1,-3)
How can you solve 2x equals 5y and 3x-2y equals 0 by graphing?
If you were to draw both graphs, then the point that they cross will give you the solutions for both x and y at the same time. To draw the graphs, get them into the form y = mx+c, where m is the gradient and c the y intercept. We have then y = 2/5x +0 & y = 3/2x+0 Therefore the only solution is x = y = 0
What is the locus of points at a given distance of a line?
The locus of points at a given distance to a line would be a line parallel to the first line. Assuming that both lines are straight.
What is the locus of points equidistant from two parallel lines?
It's a third line, parallel to both and midway between them.
What is the locus of all points in a plane equidistant from two parallel lines in the plane?
It's another line, parallel to both of the first two and midway between them.
Are the graphs of the lines in the pair parallel Explain y equals 4x plus 6 -15x plus 3y equals -45?
No because the slope in both lines have different values
How are bar graphs and line graphs similar?
They both give you info.
How are bar graphs and picture graphs alike?
Both bar graphs and picture graphs show statistics (data) in a visual (graphic) form.
How are bar graphs and circle graphs the same?
They both progression up or down
What do line graphs and bar graphs have in common?
Bar graphs and line graphs are designed to show different values of two or more subjects. They both organize data. They both use an x-axis and a y-axis.
Why can bar graphs be converted to line graphs?
They can both show the same data. You can use quantitative or categorical data with both of them.
What is the locus of points that satisfies both equations 2x-3y equals 5 and x-2y equals 1?
2x-3y=5 and x-2y=12x-3y+2y=5 and x-2y+2y=12x-y=5 and x=12(1)-y=52-y=52-y-2=5-2-y=3y=-3(x,y) is (1,-3)
How can you solve 2x equals 5y and 3x-2y equals 0 by graphing?
If you were to draw both graphs, then the point that they cross will give you the solutions for both x and y at the same time. To draw the graphs, get them into the form y = mx+c, where m is the gradient and c the y intercept. We have then y = 2/5x +0 & y = 3/2x+0 Therefore the only solution is x = y = 0
What should graphs and tables both have?
data
