Chavez Davey
06/10/2024 · Senior High School

Solve the problems. (1) The length of a rectangle is \( y \) times its width. The width of the rectangle is \( 4 y^{2} \). a. What is the area of the rectangle? b. If \( y \) is 4 , what is the area?

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To solve this problem, we need to use the given information about the rectangle's dimensions. a. To find the area of the rectangle, we use the formula for the area of a rectangle, which is length times width. Given: - The length of the rectangle is \( y \) times its width. - The width of the rectangle is \( 4y^2 \). Let's denote the width as \( w \) and the length as \( l \). We have: - \( w = 4y^2 \) - \( l = y \times w = y \times 4y^2 = 4y^3 \) Now, the area \( A \) of the rectangle is: \[ A = l \times w = 4y^3 \times 4y^2 = 16y^5 \] So, the area of the rectangle is \( 16y^5 \). b. If \( y \) is 4, we can substitute \( y \) with 4 in the area formula to find the area when \( y = 4 \). \[ A = 16y^5 \] \[ A = 16 \times 4^5 \] \[ A = 16 \times 1024 \] \[ A = 16384 \] So, when \( y = 4 \), the area of the rectangle is 16,384 square units.

Quick Answer

a. The area of the rectangle is \( 16y^5 \). b. When \( y = 4 \), the area is 16,384 square units.
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