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.
4y+yz is the correct answer
Answer is 6 you have to use distributive property to combine terms and than you simplify to get the answer<span />
Answer:
0.2
Step-by-step explanation:
19.62*0.2
Multiply by 0.2
Hope this helps plz hit the crown :D