Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

80 POINTS! NO LINKS

How much ribbon would be needed to go around a package that had a length 2x^2+3x-5/x^2+x-3 centimeters and width x^2-x-5/x^2+x-3 centimeters?


80 POINTS NO LINKS How Much Ribbon Would Be Needed To Go Around A Package That Had A Length 2x23x5x2x3 Centimeters And Width X2x5x2x3 Centimeters class=

Sagot :

The amount of ribbon needed to go around a package that had a length 2x² + 3x -5/(x² + x - 3) centimeters and width x² - x - 5 / (x² + x - 3) is

[tex]\frac{(2x^2+3x-5)(x^2-x-5)}{(x^2+x-3)^2}[/tex]

What is an equation?

An equation is an expression that shows the relationship between two or more number and variables.

The  length 2x² + 3x -5/(x² + x - 3) centimeters and width x² - x - 5 / (x² + x - 3) centimeters. Hence:

[tex]Area=\frac{2x^2+3x-5}{x^2+x-3} *\frac{x^2-x-5}{x^2+x-3} =\frac{(2x^2+3x-5)(x^2-x-5)}{(x^2+x-3)^2}[/tex]

The amount of ribbon needed to go around a package that had a length 2x² + 3x -5/(x² + x - 3) centimeters and width x² - x - 5 / (x² + x - 3) is

[tex]\frac{(2x^2+3x-5)(x^2-x-5)}{(x^2+x-3)^2}[/tex]

Find out more on equation at: https://brainly.com/question/2972832

#SPJ1