The input is the x-coordinate and the output is the y-coordinate. If the input is 4, plug 4 into the x variable to find the output, y.
y=3(4)-5
y=12-5
y=7
The output is 7.
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.
to get the slope of any line, all we need is two points off of it.
so let's see hmmmmmmm this line passes through (0, 1) and hmmmm (4, -2)
Absolute value is alwas positive
so all of them excet the 3rd one are false
|w|=0 has 1 solution
if the last one was |w|=1 then ther would be 2 solutions, -1 and 1
answer is |w|=0
Answer: B.) -4+3
Step-by-step explanation:
The model represents "-4+3" because the blue arrow (top) stops at 3 which is a positive and the red arrow (bottom) stops at -4 which is a negative.
