**Transcription for Educational Website**---### Homework Solutions**Solve the Following Problems:****17.** The area of the rectangle shown is \(18a^2b\). Write an expression for the unknown dimension.[Diagram of Rectangle]---**18.** The width of a rectangle is 3 cm less than the length. The perimeter is 42 cm. Find the area of the rectangle.```[Diagram of Rectangle, labeled with "6a" for the width and "8b" for the length]```- Diagram Explanation: - The diagram presents a generic rectangle. One side (height) is labeled \(6ab\) and the other side (length) is not labeled. ---### Solution Steps:For **Problem 17**:1. Let one dimension of the rectangle be \(x\).2. The area of a rectangle is given by \( \text{Area} = \text{Length} \times \text{Width} \).3. Set up the equation based on the given area \(18a^2b\): \[ \text{Length} \times \text{Width} = 18a^2b \]4. Solve for the unknown dimension.For **Problem 18**:1. Let the length of the rectangle be \(L\) cm.2. The width is then \(L - 3\) cm as it is 3 cm less than the length.3. The perimeter (P) of a rectangle is: \[ P = 2 \times (\text{Length} + \text{Width}) \]4. Substitute the given perimeter value and the expressions for length and width: \[ 42 = 2 \times (L + (L - 3)) \]5. Solve for \(L\): \[ 42 = 2 \times (2L - 3) \] \[ 42 = 4L - 6 \] \[ 48 = 4L \] \[ L = 12 \]6. The width \(W\) is: \[ W = L - 3 = 12 - 3 = 9 \text{ cm} \]7. Find the area (A) of the rectangle:

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The area of the rectangle shown is 18a?6. Write an expression for the unknown dimension.

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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