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Q: What is the y-intercept of the line with the equation 4x 2y 12?
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What is the slope of the line that has the equation 4x plus 2y equals 12?

what is the slope of the line that has the equation 4x+2y=12?


What is 3x plus 2y equals 8?

3x + 2y = 8 This is an equation. It could be the equation of a line.


What is the y-intercept of the line with the equation 4x plus 2y equals 12?

(0,6)


What is -x plus 2y equals 12?

It is a straight line equation that can be rearranged into slope intercept form: -x+2y = 12 2y = x+12 y = 0.5x+6 which is now in slope intercept form


What is the slope of the line whose equation is -2y equals 5x-12?

The slope (m) is -2.5 -2y = 5x-12 -2y/-2 = (5x-12)/-2 y=-2.5x + 6 or more neatly: y= 6-2.5x


Which equation represents a line that is parallel to the line whose equation is 2x 3y12?

2y= 3x+6


Solve for y if the equation is -2x - 2y equals -12?

-2x - 2y = -122x + 2y = 122y = 12 - 2xy = 6 - x


What is the point of contact between the tangent line x -2y plus 12 equals 0 and the circle x2 plus y2 -x -31 equals 0?

It works out that the tangent line x -2y +12 = 0 makes contact with the circle x^2 +y^2 -x -31 = 0 at the point (-2, 5) ------------------------------------------------- The equation of the tangent line can be rearranged to isolate x: x - 2y + 12 = 0 → x = 2y - 12 This can now be substituted for x in the equation of the circle and solved to find the value(s) of y: x² + y² - x - 31 = 0 → (2y - 12)² + y² - (2y - 12) - 31 = 0 → 4y² - 48y + 144 + y² - 2y + 12 - 31 = 0 → 5y² - 50y - 125 = 0 → y² - 10y - 25 = 0 → (y - 5)² = 0 This is a repeated root showing a single point of contact, so the line IS a tangent to the circle. → y - 5 = 0 → y = 5 Substitute into the equation for the line: → x = 2y - 12 = 2×5 - 12 = 10 - 12 = -2 → point of contact is (-2, 5)


What are the points of intersection of the straight line x -2y equals 8 with the curve of xy equals 24?

We solved the first equation for 'x': [ x = 2y + 8 ].Then we substituted it for 'x' in the second equation and rearranged: [ y2 + 4y - 12 = 0 ].Solutions of this quadratic equation are: y = 2 and -6. From which x = 12 and -4 .So the straight line intersects the hyperbola at (-4, -6) and at (12, 2) .


How do you solve this equation 3x plus 2y equals 12 solve for y?

3x + 2y = 12 ie 2y = 12 - 3x so y = 6 - 3x/2


Is 6 -6 a solution of 2y - 4x -12?

Without an equality sign the given expression can't be considered to be a straight line equation


What is the slope of the line that has the equation 4x 2y 12?

Impossible to say because the equal, the positive and/or the negative signs or instructions are missing in your question.