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
The best way to solve is by using elimination method.
20x = -58 - 2y
17x = -49 - 2y
Multiply second equation by -1
20x = -58 - 2y
-17x = 49 + 2y
Add equations.
3x = -9
Divide.
x = -3
Plug in -3 into one of the equations.
17(-3) = -49 - 2y
-51 = -49 - 2y
Add 49 to both sides.
-2 = -2y
Divide.
1 = y
So your solution is (-3, 1).
I hope this helps love! :)
Answer: 2.37
Step-by-step explanation:
Multiply each side by x to get rid of the denominator on the left side.
You get:
2.37 = 1x
1x = x
Therefore, 2.37 = x.
Also, any number divided by itself must be 1.
Answer:
The mode is 75
Step-by-step explanation: It is the one that appears most often
