The whole-number factors of 23 are 1 and 23 .
2x2-8x (two x squared minus 8 x)
x= 49/9 − 2y/9
Step-by-step explanation:
Move all terms that dont contain x to the right side and solve. (I solved for x i hope thats what you needed. Next time be a little more detailed. (: )
Answer: 648
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Explanation:
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.
