Aternary digit is either 0, 1, or 2. how many sequences of eight ternary digits containing a single 2 and a single 1 are possible?

56

Step-by-step explanation:

If there can be only one 1 and only one 2, the remaining digits must all be 0. The digits 1 and 2 can be anywhere in the 8 digits, and can be in either order.

There are 8 possible locations in the sequence for the 1, then 7 possible locations for the 2. The total number of possibilities is 8·7 = 56.

12000000, 21000000, 10200000, 20100000, ..., 00000012, 00000021.

where is the image?

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11