By definition, a function is odd if f(-x) = -f(x) for all x in the domain. The answer is choice C
Side note: a function is even if f(-x) = f(x) for all x in the domain
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 answer is C: The MAD is 4. Hope that helps
Answer: a) = 10
b) = 16
Step-by-step explanation:
a) as x = 5 so putting the value of x in given equation
5 multiplied by 2 is equal to 10
b) Putting value of x in given equation
= (5 + 3) 2
According to BODMAS solve the brackets first
= 8 x 2
= 16
Answer:
A;£8£(_("!£(£(£(('?_3!£+"('(3(£("('
