Question

a rectangle with an area of 240 in² has a length 10 inches shorter than five times its width. which equation could be used to solve for the length of the rectangle? a 5w² + 10w + 240 = 0 b 5w² - 10w + 240 = 0 c 5w² + 10w - 240 = 0 d 5w² - 10w - 240 = 0

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ inches. Then the length $l = 5w - 10$ inches.

Step2: Use area formula

The area formula for a rectangle is $A=l\times w$. Given $A = 240$ square - inches. Substitute $l$ and $A$ into the formula: $240=(5w - 10)\times w$.

Step3: Expand the equation

Expand $(5w - 10)w$ using the distributive property: $5w\times w-10\times w=5w^{2}-10w$. So the equation becomes $5w^{2}-10w = 240$.

Step4: Set the equation to standard form

Subtract 240 from both sides of the equation to get it in standard quadratic - form $ax^{2}+bx + c = 0$. We have $5w^{2}-10w-240 = 0$.

Answer:

D. $5w^{2}-10w - 240 = 0$