Without trying to list all of them, I just now thought of a way to figure out the number of different possibilities:
The total can be made from:
-- zero, 1, 2, or 3 quarters . . . 4 choices
-- zero, 1, or 2 dimes . . . . . . . 3 choices
-- zero or 1 nickel . . . . . . . . . . 2 choices
and
-- zero, 1, or 2 pennies . . . . . . 3 choices
So there are (4 x 3 x 2 x 3) = 72 different possible combinations of coins
Almost all of the possible combinations appear to be unique. I do
see one possible duplication: 1qtr is the same thing as (2dim + 1nkl).
That reduces the number somewhat, but I don't really know how to handle it.
So the number of different amounts of change is a few less than 72 .
I hope this answer is worth 5 points.
7/3 +3(2/3 -1/3)^2 =
2/3 -1/3 = 1/3 now you have
7/3 +3(1/3)^2
1/3^2 = 1/9
7/3 +3(1/9)
3*1/9 = 1/3
7/3 + 1/3 = 8/3, which can be changed to 2 2/3
Given that the original length of the baguette is 65, and for each day 15 gets cut off, we have the function
l(d) = 65 - 15d
where d is a positive integer representing the nth day. As a matter of fact, the possible vaalues for d are 0, 1, 2, 3, and 4. Since on the 5th day, there won't be enough baguette anymore. This shows that the function l(d) is not continous since only certain points satisfy the condition.
Thus, the function is l(d) = 65 - 15 where {d| 0 ≤ d ≤ 4} and it is discrete<span>.</span>
(<span>3·10</span>)·8=x<span>(<span>10·8)</span></span>
<span>30 ·8 = x ·80</span>
<span />
240=80x
x=3
The last or second sentence
