Bobby is right, because at x = -2, 3, and 5, the denominator of the function equals zero, which is by definition undefined.
You would solve using elimination.
y=1
x=-1
Answer:
5,009.206
Step-by-step explanation:
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:
O is in between m and p so it is 6
Step-by-step explanation:
In the figure below, O is between M and P, and N is the midpoint of MO. If =MP16 and =NO4, find OP.
