1. solve |x| < 13
a) {-13, 13}
b) {x|-13 < x < 13}
c) {x|x < -13 or x > 13}

2. |x| > 4
a) {-4, 4}
b) {x|-4 < x < 4}
c) {x|x < -4 or x > 4}

B) {x|-13 < x < 13}

C) {x|x < -4 or x > 4}

Step-by-step explanation:

Given in the question,

Remove the absolute value term.

If your absolute value is less than a number, then set up a three-part compound inequality that looks like this:

-(number on other side) < (quantity inside absolute value) < (number on other side)

-13 < x < 13

If your absolute value is greater than a number, then set up an "or" compound inequality that looks like this:

(quantity inside absolute value) < -(number on other side)

OR

(quantity inside absolute value) > (number on other side)

x < - 4 or x > 4

20

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!