480700. The different combinations of students that could go on the trip with a total of 25 student, but only 18 may go, is 480700.
The key to solve this problem is using the combination formula 
. This mean the number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed.
The total of students is n and the only that 18 students may go is r:
1) -2(n-6)
-2n+12
2) (5b-4)1/5
(5b-4)(1/5)
(5b)(1/5)+(-4)(1/5)
b+-4/5
3) 2/3(6y+9)
(2/3)(6y)+(2/3)(9)
4y+6
4) 4t-7t
-3t
5) -18v^2+23v^2
5v^2
6) 13q-30q
-17q
Divide by 9=7/11. Hope this help you
<span>3(a+(6x)y) was clearly multiplied out as seen by the 3a and 18xy, so the distributive property was used there. In addition, the commutative and associative properties state that you can rearrange sums, so those were used too </span><span />
The answer would be D: $162 because $120 x 35% or (0.35) = $42 so $120 + $42 gives you $162
