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What is the length of the line x -y equals 10 that spans the curve x squared plus y squared plus 4x plus 6y -40 equals 0 showing work?
If: x^2+y^2+4x+6y-40 = 0 and x-y =10 or as x = 10+y Then: (10+y)^2+y^2+4(10+y)+6y-40 = 0 Thus: 100+20y+y^2+y^2+40+4y+6y-40 = 0 Collecting like terms: 2y^2+30y+100 = 0 or as y^2+15y+50 = 0 When factored: (y+5)(y+10) = 0 So: y = -5 or y = -10 By substitution equations intersect at: (0, -10) and (5, -5) Length of line is the square root of: (5-0)^2 plus (-5--10)^2 = square root of 50 Therefore length of line is: square root of 50 or about 7.071 to three decimal places
Where are the points of intersection on the curve and line of x squared plus y squared plus 4x plus 6y minus 40 equals 0 and x equals 10 plus y?
x2+y2+4x+6y-40 = 0 and x = 10+y Substitute the second equation into the first equation: (10+y)2+y2+4(10+y)+6y-40 = 0 2y2+30y+100 = 0 Divide all terms by 2: y2+15y+50 = 0 (y+10)(y+5) = 0 => y = -10 or y = -5 Substitute the above values into the second equation to find the points of intersection: Points of intersection are: (0, -10) and (5, -5)
What are the points of intersection of x2 plus y2 plus 4x plus 6y -40 equals 0 with x -y equals 10 showing key stages of work?
If: x^2+y^2+4x+6y -40 = 0 and x -y = 10 Then by rearranging: x = 10+y and 2y^2+30y+100 = 0 Solving the above quadratic equation: y = -10 and y = -5 Points of intersection by substitution are: (0, -10) and (5, -5)
What is the answer of 6y-2y equals 40?
6y-2y=40 6-2=40/y 4=40/y y=10 6(10)-2(10)=40
What is 50 plus 40 plus 40 plus 40 plus 30 plus 30 plus 20 plus 10?
50 + 40 + 40 + 40 + 30 + 30 + 20 + 10 = 260
What are the curve and line coordinates for x square plus y square plus 4x plus 6y - 40 equals 0 and x - y equals 10?
I presume you want the points where the circle x2 + y2 + 4x + 6y - 40 = 0 and the line x - y = 10 meet.x - y = 10=> y = x - 10Substitute into circle equation:x2 + y2 + 4x + 6y - 40 = 0=> x2 + (x - 10)2 + 4x + 6(x - 10) - 40 = 0=> x2 + x2 - 20x + 100 + 4x + 6x - 60 - 40 = 0=> 2x2 - 10x = 0=> 2x(x - 5) = 0=> x = 0 or 5=> y = -10 or -5 respectivelyThe line meets the circle at the points (0, -10) and (5, -5).Another method with the same result:Equation 1: x2+y2+4x+6y-40 = 0Equation 2: x-y = 10 => x = 10+ySubstitute Equation 2 into Equation 1:(10+y)(10+y)+y2+4(10+y)+6y-40 = 0100+20y+y2+y2+40+4y+6y-40 = 0Collect like terms:2y2+30y+100 = 0 => (2y+20)(y+5) = 0y = -10 or y = -5Substitute the above values into Equation 2:When y = -10, x = 0When y = -5, x = 5Therefore the coordinates are: (0,-1) and (5,-5)
What is y squared plus 13y plus 40 equals 0?
If y2 + 13y = 40, then y ~ 15.55If y2 + 13y = 0, then y = -13Improved Answer:-y2+13y+40 = 0(y+5)(y+8) = 0Therefore: y = -5 and y = -8
What are the solutions to the simultaneous equations of x square plus y square plus 4x plus 6y minus 40 equals 0 and x minus y equals 10?
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
What are the coordinates of the straight line of y equals 5x plus 10 that meets the parabola of y equals x squared plus 4 at points A and B?
(6, 40) and (-1, 5)
Y plus 6-28 equals 40?
Y = 62
What is y if 30y equals 10 plus y?
If: 30y = 10+y Then: y = 10/29
What is 10 plus 10 plus 10 plus 10?
10 + 10 + 10 + 10 = 40
