Question

score: 0/10 answered: 0/7
question 1 \t\t\t 2 pts \t 1 \t details
the length of a rectangle is ten inches more than five times the width. the perimeter is 56 inches. find
the length and width.

the length is \t\t\t inches, and the width is \t\t\t inches.

check answer

question 2 \t\t\t 1 pt \t 1 \t details
solve.

\\(\frac{2y + 4}{4} = \frac{y}{9}\\)

y = \t\t\t

check answer

Explanation:

Question 1

Step1: Define variables

Let width be \( w \) inches. Then length \( l = 5w + 10 \) inches.

Step2: Use perimeter formula

Perimeter of rectangle: \( P = 2(l + w) \). Given \( P = 56 \), so \( 2((5w + 10) + w) = 56 \).

Step3: Simplify equation

Simplify: \( 2(6w + 10) = 56 \) → \( 12w + 20 = 56 \).

Step4: Solve for \( w \)

Subtract 20: \( 12w = 36 \) → \( w = 3 \).

Step5: Find length

\( l = 5(3) + 10 = 25 \).

Step1: Cross - multiply

Given \( \frac{2y + 4}{4}=\frac{y}{9} \), cross - multiply: \( 9(2y + 4)=4y \).

Step2: Expand left side

Expand: \( 18y+36 = 4y \).

Step3: Solve for \( y \)

Subtract \( 18y \): \( 36=- 14y \) → \( y =-\frac{36}{14}=-\frac{18}{7} \).

Answer:

The length is \( 25 \) inches, and the width is \( 3 \) inches.

Question 2