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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Step-by-step Solution
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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