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.
The series of numbers is adding 5 and then subtracting 2
20+5=25
25-2=23
23+5=28
28-2=26
and so on. Do you understand?
Hello :) ok so to find AC you would have to use sin. It would be sin70= x/4. To find AC you would have to multiply 4 by sin70 which is 3.758 or to the nearest hundredth would be 3.76. I hope I helped, if you need a more in depth explanation lmk
U think it’s 40 cuz 8 x40 equal 320 ;-; if it’s wrong my name will come to use
102+28=130 180-130=50, answer is 50°
