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
Q(3) = -2(3)+2 = -4
r(q(3)) = (-4)^2-1 = 15
No he would get back $4.56. Start with $20 -13.59-1.85=4.56
2 lb in each package
There are 3 types of nuts they all have 9 1/3 pounds in them. multiply 3 by 9 /13 to find out how many pounds of nuts there is. 3×9 1/3 That is 28 pounds. Now divide the 28 pounds of nuts into the 14 packages. 28 ÷ 14 There is 2 pounds of nuts in each package.
