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
10 times 'x' is 10 times whatever the value of 'x' is. I hope that makes sense.
The formula for depreciation is:

Where x = Initial value,
y= Amount after depreciation.
r= Rate of depreciation,
t = time (in years)
According to given problem,
x = 1040, y= 944 and t = 12 months =1 year.
So, first step is to plug in these values in the above formula, So,

944 = 1040 (1 -r)
Divide each sides by 1040.
0.907692308 =1 - r
0.907692308 - 1 = -r Subtract 1 from each sides.
-0.092307692 = -r
So, r = 0.09 or 9%.
Now plug in 0.09 in the above equation to get the depreciation equation. So,

So, 
b) To find the value of the bike after 5 months,
plug in t = 5 months= 5/12 = 0.41667 years in the above equation of depreciation.
So, 
y = 1040 * 0.961465659
y = 999.9242852
y = 1000 (Rounded to nearest integer).
Hence, the value of the bike after 5 months is $1000.
I beleive 1 is C, and 2 is B!
Answer:
52.3272727...
the 27 at the end is repeating