The absolute value of a real number $abs{x}$ is the non-negative value of $x$. It denotes the distance of that number from zero on a number line, and distance can never be negative.

Hence, for the given absolute values, we obtain
$$
begin{align*}
abs{-5}&=5\
abs{20}&=20\
abs{-15}&=15\
abs{12}&=12\
abs{-25}&=25.
end{align*}$$

Arranging these results in ascending order, we obtain
$$
5,12,15,20,25,
$$
or, as they were given in the exercise
$$
abs{-5},abs{12}, abs{-15},abs{20}, abs{-25}.$$

$abs{-5},abs{12}, abs{-15},abs{20}, abs{-25}$

The absolute value of a real number $abs{x}$ is the non-negative value of $x$. It denotes the distance of that number from zero on a number line, and distance can never be negative. Note that when having an absolute value of an expression such as $abs{a+b}$, we need to compute the value of that expression first, and then find the absolute value from the result.

*(a)* An absolute value of a number gives a non-negative value. Thus, we have
$$abs{-22}=22.$$

*(b)* Note that the first minus sign is not affected by the absolute value and an absolute value of a number gives a non-negative value. Thus, we have
$$-abs{-35}=-(35)=-35.$$

*(c)* As said, when having an absolute value of an expression such as $abs{a+b}$, we need to compute the value of that expression first, and then find the absolute value from the result. An absolute value of a number gives a non-negative value. Thus, we have
$$abs{-5-13}=abs{-18}=18.$$

*(d)* As said, when having an absolute value of an expression such as $abs{a+b}$, we need to compute the value of that expression first, and then find the absolute value from the result. An absolute value of a number gives a non-negative value. Thus, we have
$$abs{4-7}+abs{-10+2}=abs{-3}+abs{-8}=3+8=11.$$

*(e)* Note that the denominator is not affected by the absolute value and an absolute value of a number gives a non-negative value. Thus, we have
$$dfrac{abs{-8}}{-4}=dfrac{8}{-4}=-2.$$

*(f)* Note that both the denominator and numerator of the first fraction are affected by the absolute value and an absolute value of a number gives a non-negative value. For the second fraction, only denominator is affected by the absolute value function. Thus, we have
$$dfrac{abs{-22}}{abs{-11}}+dfrac{-16}{abs{-4}}=dfrac{22}{11}+dfrac{-16}{4}=2-4=-2.$$

a) $22$; b) $-35$; c) $18$; d) $11$; e) $-2$; f) $-2$

#### (a)

$$
begin{equation*}
left|x right|>3
end{equation*}
$$

#### (b)

$$
begin{equation*}
left|x right|leq8
end{equation*}
$$

#### (c)

$$
begin{equation*}
left|x right|geq1
end{equation*}
$$

#### (d)

$$
begin{equation*}
left|xright|ne5
end{equation*}
$$

These solutions are obtained on the basis of the $textbf{absolute value definition}$, and from a $textbf{number straiht line}$, which is best seen in some of the following tasks.

For each task we use $textbf{open dot}$ when the number is not included, and $textbf{solid dot}$ when we include the number in the solution.

#### (a)

$$
begin{equation*}
left|x right|leq3
end{equation*}
$$

#### (b)

$$
begin{equation*}
left|x right|>2
end{equation*}
$$

#### (c)

$$
begin{equation*}
left|x right|geq2
end{equation*}
$$

#### (d)

$$
begin{equation*}
left|x right|<4
end{equation*}
$$

Of course, when solving a task, we turn our attention to whether we $textbf{include a point or not}$ in a solution. If it is a full point, then we include that number under it, if not, the number below the point is not included in the solution.

#### (a)

We note that the graphs of these two functions coincide.
#### (b)

We have that, according to definition of absolute value is:

$$
begin{equation*}
g(x)=left|-x+8 right|=left|-(x-8) right|=left|x-8 right|=f(x)
end{equation*}
$$

So, from the previous equality, we see that $f(x)$ and $g(x)$ are the same functions, so and their graphs are the same.

We see that the graphs of functions under $a$ and $c$ differ in that the themes of graphics are displaced in 2, that is, in -2 on the x-axis, respectively.

And the graphics of the functions $b$ and $d$ differ in the fact that the themes of the graphics are moved to 2, that is, in -2 on the y-axis, respectively

So, the graph of the functions given in this task will have $textbf{topics}$ in point (-3.-4), analogous to the previous consideration, as can be seen from the graph below.

The graph of this function $textbf{will translate}$ to the value 1/2 to the left on $x$ axis, it is obtained on the basis of the zero value of what is below the absolute value, which means that the themes of graphics will be at the point (-1/2, 0) and pass through the point (0,1), we get this based on the value table, which we see from the graph that follows.

The graph of this function $textbf{will translate}$ to the value 2.5 on the $x$ axis, it is obtained on the basis of the zero value of what is below the absolute value, and for the value 3 in the positive direction on the y-axis, which means that the $textbf{themes}$ of the graphics will be at the point (2.5,3) and pass through points (1,0) and (4,0), which is obtained on the basis of the value table, which we can see on the basis of the graphics below: