The question is an illustration of combination and there are 729 potential pass codes available
<h3>How to determine the number of potential pass codes?</h3>
The given parameters are
Symbols available = 9
Length of pass code = 3
From the question, we understand that a symbol may be entered any number of times.
This means that each of the 9 available symbols can be used three times
So, the number of potential pass codes is
Passcodes = 9 * 9 * 9
Evaluate the product
Passcodes = 729
Hence, there are 729 potential pass codes available
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Answer:
12 units
Step-by-step explanation:
A=hbb/2=6·4/2= 12
Answer:
3 = 95
4 = 85
Step-by-step explanation:
ok first of all im sorry this problem stumped you
now let go
a and b are parallel so 3 would be 95
180 - 95 = 85
3 is equal to 95
4 is equal to 85
Answer:
1+1=11 2+2=22 ok na yan kuya or ate
