" If "(4x^(2)-3y^(2)):(2x^(2)+5y^(2))=12:19" हो,तो "x:y=?

If (4x^2-3y^2):(2x^2+5y^2)=12:19 then x:y=

Solve for (x - 1)^(2) and (y + 3)^(2) , 2x^(2) - 5y^(2) - x - 27y - 26 = 3(x + y + 5) and 4x^(2) - 3y^(2) - 2xy + 2x - 32y - 16 = (x - y + 4)^(2) .

Find each of the following products: (i) (4x + 5y) (4x - 5y) (ii) (3x^(2) + 2y^(2)) (3x^(2) - 2y^(2))

If 7x - 15y = 4x + y , find the value of x: y . Hence, use componendo and dividendo to find the values of : (i) (9x + 5y)/(9x - 5y) (iI) (3x^(2) + 2y^(2))/(3x^(2) - 2y^(2))

Find the radical center of the circles x^(2)+y^(2)+4x+6y=19,x^(2)+y^(2)=9,x^(2)+y^(2)-2x-4y=5

The radical centre of the circles x^(2)+y^(2)=9 , x^(2)+y^(2)-2x-2y-5=0 , x^(2)+y^(2)+4x+6y-19=0 is A) (0,0) (B) (1,1) (C) (2,2) (D) (3,3)

If A_(1),A_(2),A_(3) be the areas of circles x^(2)+y^(2)+4x+6y-19=0, x^(2)+y^(2)=9, x^(2)+y^(2)-4x-6y-12=0 respectively then