I: 12x-5y=0
II:(x+12)^2+(y-5)^2=169
with I:
12x=5y
x=(5/12)y
-> substitute x in II:
((5/12)y+12)^2+(y-5)^2=169
(25/144)y^2+10y+144+y^2-10y+25=169
(25/144)y^2+y^2+10y-10y+144+25=169
(25/144)y^2+y^2+144+25=169
(25/144)y^2+y^2+169=169
(25/144)y^2+y^2=0
y^2=0
y=0
insert into I:
12x=0
x=0
-> only intersection is at (0,0) = option B
The domain would be (4, infinity) so most likely your answer would be D.
I = PRT....for R....divide both sides by PT
I / PT = R <==
(f+g)(x)=5x-6+x^2-4x-8
A.(f+g)(x)=x^2+x-14 is correct:)
The correct answer is B. Parallel lines are lines in the same plane that never intersect.
