Answer:
A) x² + 2x + 6
B) x² + 2x - 7
C) ¼(x²+2x+1))
D) 6x²+12x+6
E) -x²-2x-1
Step-by-step explanation:
A) f(x) + 5 =x²+2x+1 + 5
= x² + 2x + 6
B) f(x)-8,=x^2+2x+1-8
= x² + 2x - 7
C) ¼f(x) = ¼(x²+2x+1)
D) 6f(x) = 6(x²+2x+1) = 6x²+12x+6
E) -f(x) = -(x²+2x+1) = -x²-2x-1
ANSWER:yes it is hopefully u get it right comment if it’s wrong
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
??? what type of question is that??
