Answered

At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

The length of a rectangular poster is 5 more inches than half its width. The area of the poster is 12 square inches. Solve for the dimensions (length and width) of the poster.

Sagot :

MrRoyal

The dimensions (length and width) of the poster are 6 inches and 2 inches

How to determine the dimensions (length and width) of the poster?

Represent the length with x and the width with y

From the question, we have the following parameters

x = 5 + 0.5y

Area, A = 12

The area of a rectangle is represented as:

A = xy

Substitute the known values in the above equation

(5 + 0.5y) * y = 12

Expand the bracket

5y + 0.5y^2 = 12

Multiply through by 2

10y + y^2 = 24

Rewrite as:

y^2 + 10y - 24 = 0

Expand

y^2 + 12y - 2y - 24 = 0

Factorize the expression

(y - 2)(y + 12) = 0

Solve for y

y = 2 or y = -12

The dimension cannot be negative.

So, we have

y = 2

Substitute y = 2 in x = 5 + 0.5y

x = 5 + 0.5 * 2

Evaluate

x = 6

Hence, the dimensions (length and width) of the poster are 6 inches and 2 inches

Read more about area at:

https://brainly.com/question/24487155

#SPJ1

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.