There are a total of 8 letters in student, with 6 different letters ( there are 2 s's and 2 t's).
First find the number of arrangements that can be made using 8 letters.
This is 8! which is:
8 x 7 x 6 x 5 x 4 3 x 2 x 1 = 40,320
Now there are 2 s's and 2 t's find the number of arrangements of those:
S = 2! = 2 x 1 = 2
T = 2! = 2 x 1 = 2
Now divide the total combinations by the product of the s and t's:
40,320 / (2*2)
= 40320 / 4
= 10,080
The answer is A. 10,080
Take all the non-zero digits and separate them from the zeros.
Then with zeros, how many are there?
Then multiply non-zero digits and zeros in powers.
3.482 * 10^6
Answer:
16
Step-by-step explanation:
Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
If you want to factor
, you could throw that into the quadratic formula with a = 1, b = 11 and c = 0, but the easier thing to do is to factor out what's common in those 2 terms. m is common, so when we factor it out:
That's the factored form.
By the Zero Product Property, either
m = 0 or m + 11 = 0.
So the 2 solutions to this are
m = 0 or m = -11
Not sure how far you need to go with this.
