Substitute 0.6 for \(\displaystyle x\) :

\(\displaystyle \left | 4x - 1.4 \right | + \left | x^{2} - 1 \right |\)

\(\displaystyle = \left | 4 \cdot 0.6 - 1.4 \right | + \left | 0.6^{2} - 1 \right |\)

\(\displaystyle = \left | 2.4 - 1.4 \right | + \left | 0.36 - 1 \right |\)

\(\displaystyle = \left | 1 \right | + \left | -0.64 \right |\)

\(\displaystyle =1 + 0.64\)

\(\displaystyle = 1.64\)

Substitute \(\displaystyle x = 0.6\).

\(\displaystyle \left |0.5 x - 0.7 \right | - \left |0.6 x - 0.4 \right |\)

\(\displaystyle = \left |0.5 \cdot 0.6 - 0.7 \right | - \left |0.6 \cdot 0.6 - 0.4 \right |\)

\(\displaystyle = \left |0.3- 0.7 \right | - \left |0.36 - 0.4 \right |\)

\(\displaystyle = \left |-0.4 \right | - \left | - 0.04 \right |\)

\(\displaystyle = 0.4 - 0.04 = 0.36\)

\(\displaystyle | x + 7 |\) is the absolute value of \(\displaystyle x+ 7\), which in turn is the sum of a number and  seven and a number. Therefore, \(\displaystyle | x + 7 |\) can be written as "the absolute value of the sum of a number and seven". Since it is equal to \(\displaystyle x - 3\), it is three less than the number, so the equation that corresponds to the sentence is 

"The absolute value of the sum of a number and seven is three less than the number."

\(\displaystyle f(x) = |3x - |x^{2}- 7|\; |\)

\(\displaystyle f(2) = |3 \cdot 2 - |2^{2}- 7|\; |\)

\(\displaystyle = |3 \cdot 2 - |4- 7|\; |\)

\(\displaystyle = |3 \cdot 2 - |-3|\; |\)

\(\displaystyle = |3 \cdot 2 -3 |\)

\(\displaystyle = |6 -3 |\)

\(\displaystyle = | 3 |\)

\(\displaystyle = 3\)

\(\displaystyle a \blacktriangledown b= \frac{a+1}{\left | a\right |+ \left | b\right |}\), or, equivalently,

\(\displaystyle a \blacktriangledown b=\left ( a+1 \right ) \div ( | a |+ | b | )\)

\(\displaystyle \frac{4}{5} \blacktriangledown \left (-\frac{4}{5} \right )=\left ( \frac{4}{5}+1 \right ) \div \left (\left| \frac{4}{5} \right |+ \left |- \frac{4}{5}\right | \right )\)

\(\displaystyle = \frac{9}{5} \div \left ( \frac{4}{5} +\frac{4}{5} \right )\)

\(\displaystyle = \frac{9}{5} \div \frac{8}{5}\)

\(\displaystyle = \frac{9}{5} \times \frac{5}{8}\)

\(\displaystyle = \frac{9}{8}\)

\(\displaystyle =1 \frac{1}{8}\)

\(\displaystyle p(x) = \frac{\left |x+2 \right |-1}{\left |x+1 \right |-2}\), or, equivalently,

\(\displaystyle p(x) =\left ( \left |x+2 \right |-1 \right ) \div \left ( \left |x+1 \right |-2 \right )\)

\(\displaystyle p\left (-1 \frac{1}{5} \right )=\left ( \left |-1 \frac{1}{5}+2 \right |-1 \right ) \div \left ( \left |-1 \frac{1}{5}+1 \right |-2 \right )\)

\(\displaystyle =\left ( \left | \frac{4}{5} \right |-1 \right ) \div \left ( \left |- \frac{1}{5} \right |-2 \right )\)

\(\displaystyle =\left ( \frac{4}{5} -1 \right ) \div \left ( \frac{1}{5} -2 \right )\)

\(\displaystyle = - \frac{1}{5} \div \left ( - \frac{9}{5} \right )\)

\(\displaystyle = \frac{1}{5} \div \frac{9}{5}\)

\(\displaystyle = \frac{1}{5} \times \frac{5}{9}\)

\(\displaystyle = \frac{1} {9}\)

\(\displaystyle a \triangleright b = \left | a- \frac{1}{2}b\right | + \left | \frac{1}{2} a+b\right |\)

\(\displaystyle \frac{1}{3} \triangleright 4 = \left | \frac{1}{3} - \frac{1}{2} \cdot 4\right | + \left | \frac{1}{2} \cdot \frac{1}{3} +4\right |\)

\(\displaystyle = \left | \frac{1}{3} - 2\right | + \left | \frac{1}{6} +4\right |\)

\(\displaystyle = \left | -1 \frac{2}{3} \right | + \left | 4 \frac{1}{6} \right |\)

\(\displaystyle = 1 \frac{2}{3} + 4 \frac{1}{6}\)

\(\displaystyle = 5 \frac{5}{6}\)

\(\displaystyle g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{3} \right |\)

\(\displaystyle g(16)= \left | \left | 1,000- \sqrt{16}\right | - 16^{3} \right |\)

\(\displaystyle = \left | \left | 1,000- 4 \right | - 16^{3} \right |\)

\(\displaystyle = \left | \left | 996 \right | - 16^{3} \right |\)

\(\displaystyle = \left | 996 - 16^{3} \right |\)

\(\displaystyle = \left | 996 - 4,096 \right |\)

\(\displaystyle = \left | -3,100 \right |\)

\(\displaystyle =3,100\)

Since this is an absolute value equation, we must set the left hand side equal to both the positive and negative versions of the right side. Therefore,

\(\displaystyle x-7 = 3\)

\(\displaystyle x-7 = -3\)

Solving the first equation, we see that \(\displaystyle x=10\)

Solving the second, we see that \(\displaystyle x=4\)

When we substitute each value back into the original equation \(\displaystyle \left | x -7\right | = 3\) , we see that they both check.

\(\displaystyle \left | 10 -7\right | = \left | 3 \right | =3\)

\(\displaystyle \left | 4 -7\right | = \left | -3 \right | =3\)

To solve this equation, we want to set \(\displaystyle 4x-2\) equal to both \(\displaystyle 14\) and \(\displaystyle -14\) and solve for \(\displaystyle x\).

Therefore:

\(\displaystyle 4x-2 = 14\)

\(\displaystyle 4x = 16\)

\(\displaystyle x=4\)

and

\(\displaystyle 4x-2=-14\)

\(\displaystyle 4x = -12\)

\(\displaystyle x=-3\)

We can check both of these answers by plugging them back into the inequality to see if the statement is true. 

\(\displaystyle |4(4) -2| = |16 -2| = |14| = 14\)

and

\(\displaystyle |4(-3)-2|=|-14| = 14\)

Both answers check, so our final answer is \(\displaystyle x=-3, 4\)