\(\displaystyle 0 \le |a|< b < |c|\), which means that \(\displaystyle b\) must be positive. 

If \(\displaystyle a\) is nonnegative, then \(\displaystyle a = |a| < b\). If \(\displaystyle a\) is negative, then it follows that \(\displaystyle a < 0 < b\). Either way, \(\displaystyle a < b\). Therefore, \(\displaystyle b\) cannot be the least. 

We now show that we cannot eliminate \(\displaystyle a\) or \(\displaystyle c\) as the least.

For example, if \(\displaystyle a = 4, b= 6, c = 10\), then \(\displaystyle a\) is the least;  we test both statements:

\(\displaystyle |a|< b < |c|\)

\(\displaystyle |4|< 6 < |10|\)

\(\displaystyle 4 < 6< 10\), which is true.

\(\displaystyle a+|b| = |c|\)

\(\displaystyle 4+|6| = |10|\)

\(\displaystyle 4+6 =10\), which is also true.

If \(\displaystyle a = 4, b= 6, c = - 10\), then \(\displaystyle c\) is the least; we test both statements:

\(\displaystyle |a|< b < |c|\)

\(\displaystyle |4|< 6 < |-10|\)

\(\displaystyle 4 < 6< 10\), which is true.

\(\displaystyle a+|b| = |c|\)

\(\displaystyle 4+|6| = |-10|\)

\(\displaystyle 4+6 =10\), which is also true.

Therefore, the correct response is \(\displaystyle a\) or \(\displaystyle c\) only.

\(\displaystyle 0 \le |a|< b\), so \(\displaystyle b\) must be positive. Therefore, since \(\displaystyle 0 \le |b| < c\), equivalently, \(\displaystyle 0 < b < c\), so \(\displaystyle c\) must be positive, and

\(\displaystyle |a|< b < c\)

If \(\displaystyle a\) is negative or zero, it is the least of the three. If \(\displaystyle a\) is positive, then the statement becomes

\(\displaystyle a < b < c\),

and \(\displaystyle a\) is still the least of the three. Therefore, \(\displaystyle c\) must be the greatest of the three.

If \(\displaystyle |x- 6| > 17\), then either \(\displaystyle x -6 > 17\) or \(\displaystyle x- 6 < -17\). Solve separately:

\(\displaystyle x -6 > 17\)

\(\displaystyle x -6 + 6 > 17 + 6\)

\(\displaystyle x> 23\)

or 

\(\displaystyle x- 6 < -17\)

\(\displaystyle x- 6 + 6< -17 + 6\)

\(\displaystyle x< -11\)

The solution set, in interval notation, is \(\displaystyle (-\infty, -11) \cup (23, \infty)\).

\(\displaystyle p = 0 \bigstar 5\)

Since \(\displaystyle 0 < 5\), evaluate

\(\displaystyle a \bigstar b = |a - b+1|\), setting  \(\displaystyle a = 0, b=5\):

\(\displaystyle 0 \bigstar 5 = |0 - 5+1|\)

\(\displaystyle 0 \bigstar 5 = |-4| = 4\)

\(\displaystyle q = 0 \bigstar 0\)

Since \(\displaystyle 0 = 0\), then select the pattern

\(\displaystyle 0 \bigstar 0 = 4\)

\(\displaystyle r=0 \bigstar( -5)\)

Since \(\displaystyle 0 > -5\), evaluate

\(\displaystyle a \bigstar b = |a + b-1|\), setting \(\displaystyle a = 0, b=-5\):

\(\displaystyle 0 \bigstar (-5) = |0 + (-5)-1|\)

\(\displaystyle 0 \bigstar (-5) = |-6| = 6\)

\(\displaystyle p = q= 4, r = 6\), so the correct choice is that \(\displaystyle p = q < r\).

\(\displaystyle 0 \le |m|< n < |p|\), which means that \(\displaystyle n\) must be positive. 

If \(\displaystyle m\) is nonnegative, then \(\displaystyle m = |m| < n\). If \(\displaystyle m\) is negative, then it follows that \(\displaystyle m < 0 < n\). Either way, \(\displaystyle m < n\). Therefore, \(\displaystyle n\) cannot be the least. 

Now examine the statemtn \(\displaystyle |m|+n = p\). If \(\displaystyle m = 0\), then \(\displaystyle n = p\) - but we are given that \(\displaystyle n\) and \(\displaystyle p\) are distinct. Therefore, \(\displaystyle m\) is nonzero, \(\displaystyle |m| > 0\), and 

\(\displaystyle |m| +n > 0 + n\)

and

\(\displaystyle p > n\).

\(\displaystyle p\) cannot be the least either.

\(\displaystyle 0 \le |a|< b\), so \(\displaystyle b\) must be positive. Therefore, since \(\displaystyle 0 \le |b| < c\), it follows that \(\displaystyle 0 < b < c\), so \(\displaystyle c\) must be positive, and

\(\displaystyle |a|< b < c\)

If \(\displaystyle a\) is negative or zero, it is the least of the three. If \(\displaystyle a\) is positive, then the statement becomes

\(\displaystyle a < b < c\),

and \(\displaystyle a\) is still the least of the three. Therefore, \(\displaystyle a\) must be the least of the three, and the correct choice is "None of the other responses is correct."

When dealing with absolute value bars, it is important to understand that whatever is inside of the absolute value bars can be negative or positive. This means that an inequality can be made.

In this particular case if \(\displaystyle |x-7| < 12\), then, equivalently, 

\(\displaystyle -12 < x-7 < 12\)

From here, isolate the variable by adding seven to each side.

\(\displaystyle -12 + 7< x-7 + 7 < 12 + 7\)

\(\displaystyle -5 < x < 19\)

In interval notation, this is \(\displaystyle (-5, 19)\).

First you plug in \(\displaystyle 7\) for \(\displaystyle x\) and squre it.  

This gives the expression \(\displaystyle 4-49\) which is equal to \(\displaystyle -45\).  

Since the equation is within the absolute value lines, you must make it the absolute value which is the amount of places the number is from zero.  

This makes your answer \(\displaystyle 45\).

The absolute value of a number is its distance from zero.  Absolute value is represented with \(\displaystyle \left | \right |\) or absolute value bars .  To solve this absolute value equation remove the absolute value bars and set the equation equal to positive and negative eleven.

\(\displaystyle x + 4 = 11\)

\(\displaystyle x + 4 - 4 = 11-4\)

\(\displaystyle x = 7\)

\(\displaystyle x + 4 = -11\)

\(\displaystyle x + 4 -4 = -11-4\)

\(\displaystyle x = -15\)

\(\displaystyle x = 7\) and \(\displaystyle x = -15\) are the two values of \(\displaystyle x\) which make this statement true.

\(\displaystyle \left |7+4 \right | = 11\)

and

\(\displaystyle \left |-15+4 \right | = 11\)

\(\displaystyle \left | -11\right |=11\)

Remember, because Absolute Value measures the distance from zero, the absolute value of a number whether negative or positive is always                    non-negative.

Absolute value measures distance from that number to the point of origin or zero.

\(\displaystyle \left | -12\right | = 12\)

However, there is a negative sign outside the Absolute value bars, which indicates the multiplication by \(\displaystyle -1\)

\(\displaystyle 12\) becomes \(\displaystyle -12\)

Therefore \(\displaystyle -\left | -12\right | = -12\) is the correct solution.