A rectangle with a width of 30cm has a perimeter from 100cm to 160cm. Graph a compound inequality that shows the possible lengths of the rectangle.
Part A: Use the number line below to graph the solution.
Part B. What type of compound inequality does your solution represent?
The Number line is from 0 to 100 in 10's

100 < perimeter < 160

Perimeter = 2(width + length)
100 < 2(width + length) < 160
50 < 30 + length < 80
20 < length < 50

This inequality shows the possible values of the length

Answer Link

calculista calculista · Answer 2 · 2019-02-23T18:16:18+01:00

Answer:

Part A) The graph in the attached figure

Part B) [tex]L\geq20\ cm[/tex] and [tex]L \leq 50\ cm[/tex]

Step-by-step explanation:

Part A)

we know that

The perimeter of a rectangle is equal to

[tex]P=2(L+W)[/tex]

we have

[tex]W=30\ cm[/tex]

and the perimeter

[tex]100\ cm \leq P\leq 160\ cm[/tex]

For [tex]P=100\ cm[/tex]

[tex]100=2(L+30)[/tex]

[tex]L=50-30=20\ cm[/tex]

so

[tex]L\geq20\ cm[/tex]

For [tex]P=160\ cm[/tex]

[tex]160=2(L+30)[/tex]

[tex]L=80-30=50\ cm[/tex]

so

[tex]L \leq 50\ cm[/tex]

The solution of the possible lengths of the rectangle is the interval

[tex][20,50][/tex]

All real numbers greater than or equal to [tex]20\ cm[/tex] and less than or equal to [tex]50\ cm[/tex]

[tex]20\ cm \leq L\leq 50\ cm[/tex]

see the graph in the attached figure

Part B)

we know that

A compound inequality contains at least two inequalities that are separated by either "and" or "or". The graph of a compound inequality with an "and" represents the intersection of the graph of the inequalities.

In this problem

The compound inequality is equal to

[tex]L\geq20\ cm[/tex] and [tex]L \leq 50\ cm[/tex]