Answer:
Ki
Step-by-step explanation:
The answer is -1. Hope this helps :)
Consider ∆JWZ and ∆JKZ
WZ~KJ (given)
<u>/</u><u> </u><u>WZJ</u>~<u>/</u><u> </u>KJZ (given)
JZ~JZ (common)
Therefore,
∆JWZ~∆JKZ by SAS congruence rule.
JW~ZK by CPCT.
Answer:
1) 3, 4
2) . 1, 2, 3, 4, 5, 6
3) 5, 6
4) . 1, 2, 5, 6
Step-by-step explanation:
Answer:
<em>its H</em>
Step-by-step explanation:
Im sure that this is the answer. The table cleary matches with my results.
Hope this helps,
-yuko
