Answer:
- 27 and 7
Step-by-step explanation:
To obtain f(- 6) substitute x = - 6 into f(x), that is
f(- 6) = 4(- 6) - 3 = - 24 - 3 = - 27
To obtain g(- 3) substitute x = - 3 into g(x), that is
g(- 3) = - 3(- 3) - 2 = 9 - 2 = 7
P = 25h + 600.
This is because even if he works no overtime, he still gets his 600.
Answer:
D
Step-by-step explanation:
The domain is the x value which in this case would be the hours of training
Answer:
The irrational number is 4.37124...
Step-by-step explanation:
This is because the number is non-terminating and non-repeating; as in the numbers don't repeat in the same order and they never end.
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.
