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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Answer:
Solve for y.
Step-by-step explanation:
6x+2y=10
6x+2y-6x=10-6x subtract 6x from both sides.
2y = -6x+10
2y/2 = -6x/2+10/2 divide each term by 2.
y = -3x+5
Hope this helps!
Answer:
14 1/6
Step-by-step explanation:
7(1/3)+ 9(2/6)= 16(2/3)
16(2/3)
16(2/3)-2(1/2)= 14(1/6)
17 I’m not sure but give it a try
