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What is the y-intercept of the line with the equation 4x 2y 12?
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∙ 9y ago
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∙ 9y ago
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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
Is point (2 1) on the line represented by the equation x 2y 5?
Without an equality sign and not knowing the plus or minus values of 2y and 5 it can't be considered to be a straight line equation.
How do you do 2x plus y equals -12 -4x-2y equals 30?
2x+y=-12-4x-2y=30 x=-y/2+30, you plug this into -12-4x-2y=30 and get -12+2y-120-2y=30 which gives 2y-2y=162. This is not possible, so the equation is unsolvable.
2y - 12x equals -6?
2y-12x=-6 is the equation of a line. It has infinite number of solutions which are all the points one the line. We can rewrite the equation as 2y=12x-6 and then divide both sides by 2.This gives us y=6x-3. From this we know the equation describes a line with y intercept -2 and slope 6.
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 the y-intercept of the line with the equation 4x plus 2y equals 12?
(0,6)
What is 3x plus 2y equals 8?
3x + 2y = 8 This is an equation. It could be the equation of a line.
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 point (1 2) on the line represented by the equation x 2y 5?
Without an equality sign and not knowing the plus or minus values of 2y and 5 it can't be considered to be a straight line equation.
Is point (2 1) on the line represented by the equation x 2y 5?
Without an equality sign and not knowing the plus or minus values of 2y and 5 it can't be considered to be a straight line equation.
