Question

If 4x^2+4y^2-20x-12y+34=0, then find the value of xy pls answer fast.

Answer 1

Answer:

4x^2+4y^2-20x-12y+34=0

Step-by-step explanation:

4x^2+4y^2-20x-12y+34=0

4x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=04x^2+4y^2-20x-12y+34=0