Hello!
In this case, there are no solutions because the lines are parallel, and the system of inequalities are opposite each other.
Hope this helps :))
Answer:
56
Step-by-step explanation:
The sum of two <em>even</em> consecutive integers is 114.
Let's let the first even integer be 2n, where n is <em>some</em> integer: it doesn't matter what n is, but if we multiply n by 2, we are certainly getting an even number. This is because any integer multiplied by 2 is even.
So, this means that the second number is 2n+2.
Their sum is 114. In other words:

Solve for n.
Combine like terms:

Subtract 2 from both sides:

Divide both sides by 4:

So, n is 28.
This means that the first number is 28(2) or 56.
And the second number is 58.
So, the smallest integer is 56.
And we're done!
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:
thousanths place
Step-by-step explanation:
Option 4
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And




Hence verified