/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 The length of a rectangular sign... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 31

The length of a rectangular sign is 3 feet longer than the width. If the sign's area is 54 square feet, find its length and width.

Short Answer

Expert verified
The length of the rectangular sign is 9 feet, and its width is 6 feet.

Step by step solution

01

Understanding the Information Provided

Given that the length of the rectangle is 3 feet longer than its width, if we let \( w \) represent the width in feet, then the length can be represented as \( w + 3 \) feet. The problem also states that the area of the rectangle is 54 square feet. Since Area A of a rectangle is given by the formula \( A = length \times width \), we can substitute the given values into the formula to get the equation: \( 54 = w \times (w + 3) \).
02

Simplifying the Equation

Expanding the product gives us: \( 54 = w^2 + 3w \). Subtracting 54 from both sides of the equation yields a quadratic equation: \( w^2 + 3w - 54 = 0 \).
03

Solving the Equation

To solve the quadratic equation we can use a method called factoring. This equation can be factored to: \( (w - 6)(w + 9) = 0\). This gives us two possible values for \( w \), \( w = 6 \) and \( w = -9 \). Considering that width cannot be a negative value, the width of the rectangle is thus 6 feet.
04

Finding the Length

To find the length, we substitute this value into the equation for the length yielding \( w + 3 = 6 + 3 = 9 \). So, length of the rectangle is 9 feet.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. A truck can be rented from Basic Rental for \(\$ 50\) per day plus \(\$ 0.20\) per mile. Continental charges \(\$ 20\) per day plus \(\$ 0.50\) per mile to rent the same truck. How many miles must be driven in a day to make the rental cost for Basic Rental a better deal than Continental's?

Rationalize the numerator. $$\frac{\sqrt{x}-\sqrt{y}}{x^{2}-y^{2}}$$

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\frac{2 x-1}{x-7}+\frac{3 x-1}{x-7}-\frac{5 x-2}{x-7}=0$$

Will help you prepare for the material covered in the next section. A telephone texting plan has a monthly fee of \(\$ 20\) with a charge of \(\$ 0.05\) per text. Write an algebraic expression that models the plan's monthly cost for \(x\) text messages.

Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. A local bank charges \(\$ 8\) per month plus 5垄 per check. The credit union charges \(\$ 2\) per month plus 8垄 per check. How many checks should be written each month to make the credit union a better deal?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.