The answer is b. There is a proof, but it is easily got by proving the other cases wrong.
First, we pay attention to the numerical coefficients of the terms in the series: 10, 21, 32, 43, 54, 65. Conclusively they form an arithmetic sequence with a common difference of 11. Thus, the next numerical coefficient is 76. Then, we pay attention to the letters which are just arrange alphabetically. The next letter ought to be G which needs to be capitalized. Thus, the answer is letter C. 76G.
Since we are trying to find the number of sequences can be made <em>without repetition</em>, we are going to use a combination.
The formula for combinations is:

is the total number of elements in the set
is the number of those elements you are desiring
Since there are 10 total digits,
in this scenario. Since we are choosing 6 digits of the 10 for our sequence,
in this scenario. Thus, we are trying to find
. This can be found as shown:

There are 210 total combinations.
The correct answer is B addition
You will add 4 to both sides then the equation will be 2x=5
First you plug everything in to get
2a=-5 (1)^2+9 you do the powers and multiply
2a=-5 (1)+9
2a=-5+9
2a=4
a=2