Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
Answer:
52%
Step-by-step explanation:
52 out of 100 is the same as .52, when you put that as a percentage, you get 52%

Keep in mind: The entire line segment (VX) is equal to 14.

Let's see how the line segment looks like first.
Line segment VX is 14 units long.
14
______________
V X
W is on the line segment somewhere, and VW is equal to 3.
3 ?
______________
V W X

We have to solve for the <em>?.</em> Let's put ? as x. So now we are solving for x.
We have to set up our equation like this:
x + 3 = 14
Since our unknown value plus 3 is equal to 14, we have to subtract 3 from 14 to get our answer.
14 - 3 = 11
3 11
______________
V W X

Answer:
log[(c^y)/(r^8)] this is m your answer
Answer:
Square pyramid
Step-by-step explanation:
The flat square on the bottom is 1 face and there are 4 triangle faces.
The flat square on the bottom has 4 vertices and there is the pointy 1 at the top that holds the 4 triangles.
Hope dis helps,
Have a great day.