Question

the length of a rectangle is 7 inches less than twice the width, w, of the rectangle.
part a
the quadratic function a(w) represents the area a, in square inches, of the rectangle for a given value of w. which function gives the area as a function of the width?
a. a(w)=7w - w²
b. a(w)=2w² - 7
c. a(w)=7w - 2w²
d. a(w)=2w² - 7w
part b
if the area of the rectangle is 60 square inches, what is the width of the rectangle?
enter your answer in the box.

Explanation:

Step1: Find the length expression

The length \( l \) is 7 inches less than twice the width \( w \). So, \( l = 2w-7 \).

Step2: Use the area formula for a rectangle

The area formula for a rectangle is \( A=l\times w \). Substitute \( l = 2w - 7 \) into the formula: \( A(w)=(2w - 7)\times w \).

Step3: Expand the expression

Using the distributive property \( a(b - c)=ab-ac \), where \( a = w \), \( b=2w \), and \( c = 7 \). We get \( A(w)=2w\times w-7\times w=2w^{2}-7w \).

Step4: Solve for the width when \( A(w) = 60 \)

Set \( 2w^{2}-7w=60 \). Rearrange to the standard quadratic form \( 2w^{2}-7w - 60=0 \).
For a quadratic equation \( ax^{2}+bx + c = 0\) (\(a = 2\), \(b=-7\), \(c=-60\)), use the quadratic formula \( w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(-7)^{2}-4\times2\times(-60)=49 + 480=529\).
Then \( w=\frac{-(-7)\pm\sqrt{529}}{2\times2}=\frac{7\pm23}{4}\).
We have two solutions: \( w=\frac{7 + 23}{4}=\frac{30}{4}=7.5\) and \( w=\frac{7-23}{4}=\frac{-16}{4}=- 4\). Since width cannot be negative, we discard \( w=-4 \).

Answer:

Part A: D. \( A(w)=2w^{2}-7w \)
Part B: \( 7.5 \)