Question

b.
If x+y+z= 12 and xy + yz + x2 = 47, find x'+y+z?​

Answer 1

Step-by-step explanation:

Hope this will help you

Answer 2

Answer:

x+y+z =12 xy+yz+xz = 47

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

(12) ^2 = x^2+y^2+z^2 + 2*47

144 = x^2+y^2+z^2 + 94

x^2+y^2+z^2 = 144 - 94

x^2+y^2+z^2 = 50

hope it helps you...