\(\displaystyle a \ast b = \left | \frac{1}{4}a - \frac{1}{7}b \right |\)

\(\displaystyle 7 \ast 4= \left | \frac{1}{4} \cdot 7 - \frac{1}{7} \cdot 4 \right |\)

\(\displaystyle = \left | \frac{7}{4} - \frac{4}{7} \right |\)

\(\displaystyle = \left | \frac{49}{28} - \frac{16}{28} \right |\)

\(\displaystyle = \left | \frac{33}{28} \right |\)

\(\displaystyle = \left |1 \frac{5}{28} \right |\)

\(\displaystyle = 1 \frac{5}{28}\)

\(\displaystyle a \ast b = |a+b| + |a-b|\)

\(\displaystyle \left (-7 \right ) \ast \left (-5 \right )= |-7+\left (-5 \right )| + |-7-\left (-5 \right )|\)

\(\displaystyle = |-12| + |-2|\)

\(\displaystyle = 12+2\)

\(\displaystyle = 14\)

\(\displaystyle a \ast b = |a+b| - |a-b|\)

\(\displaystyle \left (-7 \right ) \ast \left (-9 \right )= |\left (-7 \right )+\left (-9 \right )| - |\left (-7 \right )-\left (-9 \right )|\)

\(\displaystyle = |-16| - |2|\)

\(\displaystyle = 16 - 2\)

\(\displaystyle = 14\)

An absolute value of a number must always assume a nonnegative value, so

\(\displaystyle |3x- 12 | \ge 0\), and

\(\displaystyle |3x- 12 | + 5\ge 0 +5\)

Therefore, 

\(\displaystyle f(x)=|3x- 12 | + 5\ge 5\)

and the range of \(\displaystyle f\) is the set \(\displaystyle [5, \infty)\).

Before we apply the absolute value to the two terms in the equation, we simplify what's inside of them first:

\(\displaystyle \left | -4+3(-5)-(-6)\right |+3x=\left |7-2(4) \right |\)

\(\displaystyle \left | -4-15+6\right |+3x=\left |7-8 \right |\)

\(\displaystyle \left | -13\right |+3x=\left |-1 \right |\)

Now we can apply the absolute value to each term. Remember that taking the absolute value of a quantity results in solely its value, regardless of what its sign was before the absolute value was taken. This means that that absolute value of a number is always positive:

\(\displaystyle \left | -13\right |+3x=\left |-1 \right |\)

\(\displaystyle 13+3x=1\)

\(\displaystyle 3x=-12\)

\(\displaystyle x=-4\)

The key to answering this question is to note that this equation can be rewritten in piecewise fashion. 

If \(\displaystyle x \ge 10\), since both \(\displaystyle x-10\) and \(\displaystyle x-6\) are nonnegative, we can rewrite \(\displaystyle f\) as

\(\displaystyle f(x)= x-10 + x-6\), or

\(\displaystyle f(x) = 2x-16\).

On  \(\displaystyle [10, \infty)\), this has as its graph a line with positive slope, so it is an increasing function. The range of this part of the function is \(\displaystyle [f(10), \infty)\), or, since

\(\displaystyle f(10)= |10-10|+|10-6| = 0 + 4 = 4\), 

\(\displaystyle [4, \infty)\).

If \(\displaystyle 6 < x < 10\), since \(\displaystyle x-10\) is negative and \(\displaystyle x-6\) is positive, we can rewrite \(\displaystyle f\) as

\(\displaystyle f(x)= -x+10 + x-6\), or

\(\displaystyle f(x)=4\)

\(\displaystyle f\) is a constant function on this interval and its range is \(\displaystyle \{4 \}\).

If \(\displaystyle x \le 6\), since both \(\displaystyle x-10\) and \(\displaystyle x-6\) are nonpositive, we can rewrite \(\displaystyle f\) as

\(\displaystyle f(x)= -x+10 -x+6\), or

\(\displaystyle f(x) = -2x+16\).

On  \(\displaystyle [- \infty, 6)\), this has as its graph a line with negative slope, so it is a decreasing function. The range of this part of the function is \(\displaystyle [f(6), \infty)\), or, since 

\(\displaystyle f(6)= |6-10|+|6-6| = 4+0=4\), \(\displaystyle [4, \infty)\).

The union of the ranges is the range of the function - \(\displaystyle [4, \infty)\).

The key to answering this question is to note that this equation can be rewritten in piecewise fashion. 

If \(\displaystyle x > 10\), since both \(\displaystyle x-10\) and \(\displaystyle x-6\) are positive, we can rewrite \(\displaystyle f\) as

\(\displaystyle f(x)= x-10 - x+6\), or

\(\displaystyle f(x)= -4\),

a constant function with range \(\displaystyle \left \{ -4\right \}\).

If \(\displaystyle 6 \le x \le 10\), since \(\displaystyle x-10\) is negative and \(\displaystyle x-6\) is positive, we can rewrite \(\displaystyle f\) as

\(\displaystyle f(x)=-x+10-x+6\), or

\(\displaystyle f(x)=-2x+16\)

This is decreasing, as its graph is a line with negative slope. The range is \(\displaystyle [f(10), f(6)]\),

or, since

\(\displaystyle f(10)= |10-10|- |10-6| = 0 - 4 = -4\)

and

\(\displaystyle f(6)= |6-10|- |6-6| = 4 - 0 = 4\),

\(\displaystyle [-4,4]\).

If \(\displaystyle x < 6\), since both \(\displaystyle x-10\) and \(\displaystyle x-6\) are negative, we can rewrite \(\displaystyle f\) as

\(\displaystyle f(x)= -x+10 +x-6\), or

\(\displaystyle f(x)= 4\),

a constant function with range \(\displaystyle \left \{ 4\right \}\).

The union of the ranges is the range of the function - \(\displaystyle [-4,4]\) - which is not among the choices.

This question plays a few tricks dealing with absolute values. To begin, we can recognize that any negative sign within an absolute value can basically be rendered positive. So this:

\(\displaystyle \tiny \tiny \small \left | -34\right |*\left | -2\right |*-\left | -0.5\right |\)

becomes:

\(\displaystyle \small 34*2*-\left | 0.5\right |\)

In this case, we still have a negative that was outside of the absolute value sign. This term will stay negative, so we get:

\(\displaystyle \small \small 34*2*- 0.5=-34\) 

This makes our answer \(\displaystyle \small -34\).

To solve this absolute value inequality, we must remember that the absolute value of a function that is less than a certain number must be greater than the negative of that number. Using this knowledge, we write the inequality as follows, and then perform some algebra to solve for \(\displaystyle x\):

\(\displaystyle \left | 2x+3\right |< 6\)

\(\displaystyle -6< 2x+3< 6\)

\(\displaystyle -9< 2x< 3\)

\(\displaystyle -\frac{9}{2}< x< \frac{3}{2}\)

To solve this absolute value inequality, we must remember that the absolute value of a function that is greater than a certain number is also less than the negative of that number. With this in mind, we rewrite the inequality as follows and then solve for the possible intervals of \(\displaystyle x\):

\(\displaystyle \left | 2x-3\right |>5\)

\(\displaystyle 2x-3< -5\)   or   \(\displaystyle 2x-3>5\)

\(\displaystyle 2x< -2\)   or   \(\displaystyle 2x>8\)

\(\displaystyle x< -1\)   or   \(\displaystyle x>4\)