(pi)(8y^(2)+6y+1)#(2y+1)

Factorise: 12(y^(2)+7y)^(2)-8(y^(2)+7y)(2y-1)-15(2y-1)^(2)

x ^ (2) + y ^ (2) + 1, x ^ (2) + 2y ^ (2) + 3, x ^ (2) + 3y ^ (2) + 4y ^ (2) + 2,2y ^ (2) + 6.3y ^ (2) + 8y ^ (2) + 1.2y ^ (2) + 3.3y ^ (2) +4] |

(i) 15x^(2) - 8x + 1 = 0 (ii) 6y^(2) - 19y + 10 = 0

The area of the region x+y<=6,x^(2)+y^(2)<=6y and y^(2)<=8y is

int_(0)^(4)((y^(2)-4y+5)sin(y-2)dy)/([2y^(2)-8y+1]) is equal to

int_(0)^(4)((y^(2)-4y+5)sin(y-2)dy)/([2y^(2)-8y+1]) is equal to

int_(0)^(4)((y^(2)-4y+5)sin(y-2)dy)/([2y^(2)-8y+1]) is equal to

Show that the circles x^(2) + y^(2) + 6x + 2y + 8 = 0 and x^(2) + y^(2) + 2x + 6y + 1 = 0 intersect each other.

Divide 8y^(3)-6y^(2)+4y-1 by 4y+2. Also write the quotient and the remainder.