Answer:
the missing numbers be 60 and 6 respectively
Step-by-step explanation:
The computation of the missing numbers are shown below:
Let us assume the missing numbers be x
a/
, 
b.

Hence, the missing numbers be 60 and 6 respectively
Answer:
I think the answer is 19
Step-by-step explanation:
Answer:

Step-by-step explanation:
Answer:
C is false
Step-by-step explanation:
Example 3 is odd but 6 is equal.
3•6 is 18
18 is even