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A rectangle is 2x-3 in by 3x-1 in. Find the area.

2 Answers

6 votes

Answer:


\huge\boxed{A=6x^2-11x+3;\ \text{for}\ x>1.5}

Explanation:

The formula of an area of a rectangle:


A=a\cdot b\\\\a,\ b-\text{sides of a rectangle}

We have:


a=2x-3\\b=3x-1

The domain:


2x-3>0\ \wedge\ 3x-1>0\\\\2x-3+3>0+3\ \wedge\ 3x-1+1>0+1\\\\2x>3\ \wedge\ 3x>1\\\\(2x)/(2)>(3)/(2)\ \wedge\ (3x)/(3)>(1)/(3)\\\\x>1.5\ \wedge\ x>(1)/(3)\Rightarrow\boxed{x>1.5}

Substitute:


A=(2x-3)(3x-1)

use FOIL: (a + b)(c + d) = ac + ad + bc + bd


A= (2x)(3x)+(2x)(-1)+(-3)(3x)+(-3)(-1)\\\\A=6x^2-2x-9x+3

combine like terms


A= 6x^2+(-2x-9x)+3\\\\A=6x^2-11x+3

User Harrakiss
by
8.8k points
2 votes

Answer:

6x² - 11x + 3

Explanation:

To find the area of a rectangle, we multiply the length by the width.

A = l • w

length: 2x-3

width: 3x-1

A = (2x-3) • (3x-1)

To multiply these two binomials, we must FOIL (first | outer | inner | last) them.

F: 2x • 3x = 6x²

O: 2x • -1 = -2x

I: -3 • 3x = -9x

L: -3 • -1 = 3

Combine the terms.

6x² - 2x - 9x + 3

Combine like terms.

6x² - 11x + 3

This is your area.

Hope this helps!

User Slowyn
by
7.9k points

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