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 · 2019-02-23T12:16:18-05:00

Answer:

Part A) The graph in the attached figure

Part B) \(L\geq20\ cm\) and \(L \leq 50\ cm\)

Step-by-step explanation:

Part A)

we know that

The perimeter of a rectangle is equal to

\(P=2(L+W)\)

we have

\(W=30\ cm\)

and the perimeter

\(100\ cm \leq P\leq 160\ cm\)

For \(P=100\ cm\)

\(100=2(L+30)\)

\(L=50-30=20\ cm\)

so

\(L\geq20\ cm\)

For \(P=160\ cm\)

\(160=2(L+30)\)

\(L=80-30=50\ cm\)

so

\(L \leq 50\ cm\)

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

\([20,50]\)

All real numbers greater than or equal to \(20\ cm\) and less than or equal to \(50\ cm\)

\(20\ cm \leq L\leq 50\ cm\)

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

\(L\geq20\ cm\) and \(L \leq 50\ cm\)