Answer:
b. y = -9
Step-by-step explanation:
A line with zero slope has no dependence on x. It is simply ...
... y = constant
The constant must be the same as the y-value of the point you want the line to go through. Since the point is (12, -9), the constant is -9 and the equation is ...
... y = -9
Answer:
The slope is 2
Step-by-step explanation:
Pick two points and plug into this equation: y2-y1/x2-x1
(-3,2) and (-1,6)
2-6/-3-(-1)=-4/-2= 2
Explanation + answers
Because there are lengths going from 0 to 1, the lines must mean either decimals or fractions (we'll use fractions for this.)
1. There are twelve lines from 0 and 1, which we can use as the denominator for our fraction. This means the length of each line is 1/12.
2. In order to find where K's point is at, we simply need to count until we get to it. After counting, I see that K is on point 8/12, which we can simplify to get a smaller number. If we simplify once, we get 4/6, which we can again simplify to get 2/3. This gives us the answer K is on point 8/12 or 2/3.
Answer:
A) JK and AB =
![[tex] \boxed{( \theta) = \f{12}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%5Cf%7B12%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, =》 
=》 
[/tex]
B) AABC
![[tex] \boxed{( \theta) = {1}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%7B1%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, 
C) AABM:
![[tex] e \fbox{: }](https://tex.z-dn.net/?f=%5Btex%5D%20e%20%5Cfbox%7B%3A%20%7D)
The value of k for which 
is 
ans[/tex]
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