There are as many palindromic 12-bit binary strings as there are permutations with repetition of 6-bit binary strings, which is equal to
.
There are also only two permutations with repetition of 6-bit binary strings, that when each of them is mirrored and the result is "glued" to the right side of the original string, will give a palindromic string without 10 as a substring.
Those strings are 000000 and 111111.
Therefore there are
palindromic 12-bit binary strings with 10 as substring.
Answer:
Step-by-step explanation:
This equation is solved if one of the following 2 conditions is met
1.
- (x² + 1) / 10 is between 1 and 2
and
- 10 / (x² + 1) is between 0 and 1
<u>Solve this:</u>
- 1 < (x² + 1) / 10 < 2
- 10 < x² + 1 < 20
- 9 < x² < 19
- 3 < |x| < √19
- x ∈ (- √19, - 3) ∪ (3, √19)
- 0 < 10 / (x² + 1) < 1
- x² + 1 > 10
- x² > 9
- |x| > 3
Solution for this case is x ∈ (- √19, - 3) ∪ (3, √19)
2.
- (x² + 1) / 10 is between 0 and 1
and
- 10 / (x² + 1) is between 1 and 2
<u>Solve this:</u>
- 0 < (x² + 1) / 10 < 1
- 0 < x² + 1 < 10
- - 1 < x² < 9
- 0 ≤ |x| < 3
- x ∈ ( - 3, 3)
- 1 < 10 / (x² + 1) < 2
- x² + 1 < 10 ⇒ x² < 9 ⇒ |x| < 3
- x² + 1 > 5 ⇒ x² > 5 ⇒ |x| > √5
Solution for this case is x ∈ (- 3, - √5) ∪ (√5, 3)
Answer:
Correct amnswer is A
Step-by-step explanation:
Anna= a
Vickie= v
Luanna= l
a is 3 years older that v
l is 2 years younger than v
Anna's age; 3+v
Luanna's age; v-2