Real solution in this system of equations
BRAINLIEST WILL BE GIVEN

A

Step-by-step explanation:

Firstly we find the x value. x-7=0; x=7

Secondly we introduce x value in 1st ecuation. So

(7-3)^2 +(y+1)^2=16, 16+y^2+2y+1=16,

Consider this ecuation: y^2+2y+1=0

y1=( -b+Δ)/ 2a, y2= (-b-√Δ)/2a, Δ=√b^2-4ac=√4-4=0

y1,2= -2/2= -1  

1 Real solution x=7 and y= -1.

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation: