18 times lol !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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
Figure this out as though no digits are present. You can choose 8 of the 26 letters and that's it.
26^8 [you can allow repeating letters] = 2.088 * 10^11
The total number of ways that can be chosen with letters and digits intermixed = 36^8
The probability of getting no numbers is 26^8 / 36^8 or (26/36)^8 = 0.07402
So the probability that there were be at least one digit is
1 - 0.07402 = 0.92598
Hey there Caleb!
To start of, the sentence was clear on that it said . . .

The statement listed above is the (key) point. We need to keep this in mind.
So, now that we know this.
We would use the mathematic (Ф)
So,sense we know that a (hexagon) had it's (6) shapes, we would do
. . .
Ф = 360° : 6 = 60°
Your correct answer to this question would be
. . .

Hope this helps.
~Jurgen
The team gained 3 yards on the play audition missed