20x+12y = 2012 12y = -20x+2012 y = -5/3x+503/3 Therefore the perpendicular slope is the positive reciprocal -5/3 which is 3/5
Improved Answer:-If: 20x+27 = 143Then: x = 5.8
280
y + x = 9 10y + 20x = 100 This is a basic linear system of equalities. We can start by rewriting the first equation as: -10y - 10x = -90 by multiplying the equation by -10, from here we can eliminate y and solve for x (-10y - 10x = -90) + (10y + 20x = 100) = (10x = 10) This yields x = 1, and by plugging x into the first equation y = 8
2x + 20x = 111,45222x = 111,452x = 5,066
20x+12y = 2012 12y = -20x+2012 y = -5/3x+503/3 Therefore the perpendicular slope is the positive reciprocal -5/3 which is 3/5
If: 20x+80y = 0 Then: 80y = -20x And: y = -0.25x So the slope is -0.25 and it has no y intercept
Improved Answer:-If: 20x+27 = 143Then: x = 5.8
Oh, dude, you just gotta rearrange that equation a bit. So, like, first divide by 5 to get -4x - y = 8. Then, if you wanna be all fancy and use slope-intercept form, just solve for y to get y = -4x + 8. And there you have it, a technically correct answer with a sprinkle of my signature humor.
20x + 80y = 0Subtract 20x from each side of the equation:80y = -20xDivide each side by 80:y = -1/4 xm = -1/4b = 0
2
2x + 5 = 25 2x = 20x = 10
280
NO,because in quadratic equation highest power of a variable is 2. e.x. 4x2-20x+10=0.
y + x = 9 10y + 20x = 100 This is a basic linear system of equalities. We can start by rewriting the first equation as: -10y - 10x = -90 by multiplying the equation by -10, from here we can eliminate y and solve for x (-10y - 10x = -90) + (10y + 20x = 100) = (10x = 10) This yields x = 1, and by plugging x into the first equation y = 8
2x + 20x = 111,45222x = 111,452x = 5,066
If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)