Vani has some nickels and some dimes. She has a minimum of 17 coins worth no more than $1.40 combined. If Vani has 9 nickels, determine all possible values for the number of dimes that she could have.
Let's assume that Vani has $x worth of dimes. Therefore, a dime is $0.10 so she has 0.1x dimes.
Since Vani has 9 nickels, each which is worth $0.05, then she has 9*0.05 = $<<9*0.05=0.45>>0.45 worth of nickels.
If Vani has a minimum of 17 coins worth no more than $1.40 combined, we can write the inequality 0.1x + 0.45 ≤ 1.40.
Combining like terms, we get 0.55 + 0.1x ≤ 1.40
Subtracting 0.55 from both sides of the inequality, we get 0.55 - 0.55 + 0.1x ≤ 1.40 - 0.55
0.1x ≤ 0.85
Dividing both sides of the inequality by 0.1, we get x ≤ <<0.85/0.1=8.5>>8.5
Since we cannot have a fraction of a dime, the possible values for the number of dimes that Vani could have are 0, 1, 2, ..., 8.
Therefore, Vani could have a minimum of 17 - 9 = <<17-9=8>>8 dimes. Answer: \boxed{8}.
wrong
I apologize for the mistake in my previous response. Let's revise the solution.
Let n be the number of nickels and d be the number of dimes that Vani has.
From the given information, we know that Vani has at least 17 coins. Therefore, we have the inequality:
n + d ≥ 17
The value of the nickels is 5n cents, and the value of the dimes is 10d cents. From the given information, we know that the combined value of all the coins is no more than $1.40 or 140 cents. Therefore, we have the inequality:
5n + 10d ≤ 140
Now, we can solve the system of inequalities:
n + d ≥ 17
5n + 10d ≤ 140
To determine the possible values for the number of dimes, we can substitute the minimum value for n, which is 9 (since Vani has 9 nickels), into the second inequality:
5(9) + 10d ≤ 140
45 + 10d ≤ 140
10d ≤ 140 - 45
10d ≤ 95
d ≤ 9.5
Since the number of dimes must be a whole number, the possible values for the number of dimes that Vani could have are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
Therefore, the possible values for the number of dimes is 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
Answer: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.