We have this set of digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} From that set, we can only pick three items. We cannot select the same digit twice.
Consider a blank three digit number such that it is composed of slot A, slot B, slot C.
Since the number must be larger than 100, this means that we cannot select 0 as the first digit. We go from a pool of 10 digits to 10-1 = 9 digits for our first selection. In other words, we have this subset to select from {1, 2, 3, 4, 5, 6, 7, 8, 9} So we have 9 choices for slot A.
For slot B, we also have 9 choices since 0 is now included. For instance, if we selected the digit '4' then we have this subset of choices left over: {0, 1, 2, 3, 5, 6, 7, 8, 9} which is exactly 9 items.
For slot C, we have 9-1 = 8 items left to choose from. For example: If we choose '4' for slot A, and '2' for slot B, then we have this subset to choose from: {0, 1, 3, 5, 6, 7, 8, 9} exactly 8 items
In summary so far, we have... 9 choices for slot A 9 choices for slot B 8 choices for slot C
Giving a total of 9*9*8=81*8=648 different three digit numbers. You'll notice that I'm using the counting principle which allows for the multiplication to happen. Think of a probability tree.
The functional equation of the line is f (x)=a*x+b you have already a: -4 -->f (x)=-4 x+b Now you can use the point (3;-3) 3 is the x-coordinate and -3 the f (x)-coordinate --> -3 = -4*(3) + b Now you can solve it -3 = -12+b (+12) b = 9 ANSWER: f (x) = -4x + 9
