Kerel is creating a rectangular dog run in his backyard. The length of the dog run is

Mathematics

2 answers:

Rama09 [41]2 years ago

8 0

The range of values for the width of the dog run is 73.5 ≥ w ≤ 237

<h3>Compound inequality</h3>

Perimeter of a rectangle = 2(length + width)

Length = 67 feet
Width = w

The inequality:

281 ≥ 2(67 + w) ≤ 608

281 ≥ 134 + 2w ≤ 608

solve independently

281 ≥ 134 + 2w

281 - 134 ≥ 2w

147 ≥ 2w

w ≥ 147/2

w ≥ 73.5

2(67 + w) ≤ 608

134 + 2w ≤ 608

2w ≤ 608 - 134

2w ≤ 474

w ≤ 474 / 2

w ≤ 237

Learn more about inequality:

brainly.com/question/25275758

#SPJ1

Hunter-Best [27]2 years ago

3 0

The width of the dug run is 73.5 ≤ w ≤ 237

<h3>What is a compound inequality?</h3>

A compound inequality is an inequality formed by combining two simple inequalities.

Analysis:

Let the width of dug run be w

perimeter = 2(67 + w) = 134 + 2w

281 ≤ 134 + 2w

147 ≤ 2w

73.5 ≤ w

Also, 134 + 2w ≤ 608

2w ≤ 474, w ≤237

In conclusion, the range of values of the width of the dug run is 73.5≤w≤237

Learn more about compound inequality: brainly.com/question/1604153

#SPJ1

You might be interested in

Plz help with this question.

Galina-37 [17]

Answer:

$Both \: the \: moon \: and \: earth \: are \: in \: \\ the \: shape \: of \: spheres. \\ Surface\: area \: of \: a \: sphere \: of \: radius \: 'r' \\$

Let d1 be the diameter of the moon and d2 and be the diameter of the earth.Let r1 be the radius of the moon and r2 be the radius of the earth.

$given \: d1 = \frac{1}{4}d2 \\ = > 2r1 = \frac{1}{4} ×2r2 \\ =>r1 = \frac{1}{4}×r2 \\ Now, \: ratio \: of \: \: their \: surface \: areas \: is: \\ = s1: s2 \\ = r {1}^{2}: r {2}^{2} \\ = {1}^{2 } : {4}^{2 } \\ =1:16$

<h2> hope it helps you</h2>