Answer:
sry I just wanted the points I'm in middle school so I don't know this stuff either
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
Answer:
im preety sure its c
Step-by-step explanation:
Does not exist, since the graph of cosine is continuous between the intervals [-1,1]. You do not know where it will be when it “reaches” infinity
Answer:
Here is the ans..hope it helps:)
