Question

if x+y+z=12&1 x2+y2+ z2=64, then
find the xy + yz + xz​

Answer 1

Answer:

xy+yz+zx=40

Step-by-step explanation:

(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)

12^2=64+2(xy+yz+zx)

2(xy+yz+zx)=144-64

xy+yz+zx=80/2

xy+yz+zx=40