4y^(2)+3y+5=0

Subtract : 5y^(4) - 3y^(3) + 2y^(2) + y - 1 from 4y^(4) - 2y^(3) - 6y^(2) - y + 5

The equation to the pair of lines passing through (1, 1) and parallel to the pair of lines 2x^(2)-3xy+4y^(2)+3x-5y+2 = 0 is

Solution of D.E (dy)/(dx)=(2x+5y)/(2y-5x+3) is,if (y(0)=0) (1) x^(2)-y^(2)+5xy-3y=0 (2) x^(2)+y^(2)+5xy-3y=0 (3) x^(2)-y^(2)+5xy+3y=0 (4) x^(2)-y^(2)-5xy-3y=0

The lines parallel to the normal to the curve x y=1 is/are 3x+4y+5=0 (b) 3x-4y+5=0 4x+3y+5=0 (d) 3y-4x+5=0

The lines parallel to the normal to the curve x y=1 is/are (a) 3x+4y+5=0 (b) 3x-4y+5=0 (c) 4x+3y+5=0 (d) 3y-4x+5=0

The lines parallel to the normal to the curve x y=1 is/are (a) 3x+4y+5=0 (b) 3x-4y+5=0 (c) 4x+3y+5=0 (d) 3y-4x+5=0

The lines parallel to the normal to the curve x y=1 is/are (a) 3x+4y+5=0 (b) 3x-4y+5=0 (c) 4x+3y+5=0 (d) 3y-4x+5=0

the radical axis of the circles x^(2) + y^(2) + 3x + 4y - 5 = 0 " and " x^(2) + y^(2) - 5x + 5y - 6 = 0 is

the slope of the radical axis of the circles x^(2) + y^(2) + 3x + 4y - 5 = 0 " and " x^(2) + y^(2) - 5x + 5y - 6 = 0 is

The radical axis of circles x^(2) + y^(2) + 5x + 4y - 5 = 0 and x^(2) + y^(2) - 3x + 5y - 6 = 0 is