Is 102 divisible by 4 5 6 9 or 10 i am failing and about to

It can't be divided by 5 or 10 since when you count by 10's the numbers end in only zeros and when you count by 5 it is either 5 or 0  (5, 10, 15, 20 and so on).

9 will only work if the digits in the answer add up to 9.  18: 1+8=9
99: 9+9=18 and 1+8=9  So it can't be divided by 9  102: 1+0+2=3 (not 9)

That leaves 6 and six works - here's why.
There is a rule for 3's that says that the sum of the digit must add up to 3, 6 or 9.
Look:
3x1 = 3
3 x 2 = 6
3 x 3 = 9
3 x 4 = 12 and 1+2=3
3 x 5 = 15 and 1+ 5 = 6
3 x 6 = 18 and 1+ 8 = 9  
3 x 7 = 21  and 2 +1 = 3  and so on -  do you see the 3,6,9 pattern repeating.

Six follows the same rule except the number must be even.  Look above to see that 6 works with 6, 12 and 18 but not the others.  
So 102 is divisible by 3  (1+0+2 = 3)  AND it is an even number - therefore it is divisible by 6.  Here's proof: 6 x 15 = 90 and 102 is 12 more than 90 and six goes into that 12)

I hope that this will help you feel better. I don't want you to cry. :-)

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.