Answer with Step-by-step explanation:
We are given that A,B,C,D,E and F.
We have to find the number of different four -letter arrangements can be formed using given six letters a,if the first letter must be C and one of the other letters must be B and no letter can be used more than once in the arrangement.
Number of letters=6
We have to arrange four letter out of six
After fixing C and B then we choose only two letters out of remaining four letters and repetition is not allowed.
Permutation formula :
We have n=4 an r=2
Using this formula and substitute the values
Then, we get 

Hence, number of different four -letter arrangements can be formed using six letters when repetition is not allowed=12
I'm pretty sure this answer is A or D
If not then I don't know it but I'll try to help you out
We take the equation
<span>x^2 -4x + y^2 + 6y + 12 = 0
and complete the square for x and y.
</span><span>x^2 -4x + 4 - 4 + y^2 + 6y + 9 - 9 + 12= 0</span>
<span> (x-2)^2 - 4 + (y+3)^2 - 9 = -12</span>
<span> <span>(x-2)^2+ (y+3)^2 = 1
Therefore the answer is </span></span><span>(x-2)^2+ (y+3)^2 = 1</span>
Answer:
A
Step-by-step explanation:
Answer:
8$
Step-by-step explanation:
32 divided by 4