We cant Answer it because the amount of cards presented is incorrect
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.
6) subtract 16 on both sides
then it should be x/4>18
then you would times 18 by 4 and get
c as the answer
7) the answer is d.
because 52-12 is 40 and 52/4 is 13
Answer:
no no
Step-by-step explanation:
nah nahnah nahnah nahnah nahnah nahnah nahnah nah
Answer:
A) 2k+4
Step-by-step explanation:
To solve for terms of x, we need to. she sure that x is on its own side while every other term is on the other side.
To start, we will cross multiply. If we have a/b=c/d, then ad=bc
We can perform the same calculation here,
6/x=3/(k+2)
6(k+2)=3x
Divide both sides by 3
2(k+2)=x
x=2k+4
