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The length of a rectangle is 1 ft more than twice the width, and the area of the rectangle is 45 ft^2. Find the dimensions of the rectangle.

[tex]\text{Width} =x\\\\\text{Length}=2x+1\\\\~~~~~~~\text{Area} = 45~\text{ft}^2\\\\\implies x(2x+1) =45\\ \\\implies 2x^2+x = 45\\\\\implies 2x^2 +x-45=0\\\\\implies 2x^2+10x -9x -45 =0\\\\\implies 2x(x+5) -9(x+5)=0\\\\\implies (2x-9)(x+5) =0\\\\\implies x=\dfrac 92, ~~ x\neq -5\\\\\text{Hence,}\\\\\text{Width} = \dfrac 92 ~ \text{ft}\\\\\text{Length} = 2\cdot \dfrac 92 +1 = 10~ \text{ft}[/tex]

Аноним Аноним · Answer 2 · 2022-04-18T15:00:15+02:00

Аноним Аноним

18-04-2022

Answer:

Length 10 ft.

Width 4.5 ft.

Step-by-step explanation:

If the width is x ft then the length is 2x + 1 ft.

Area = width * length, so:

45 = x(2x + 1)

45 = 2x^2 + x

2x^2 + x - 45 = 0

2x^2 + 10x - 9x - 45 = 0

2x(x + 5) - 9(x + 5) = 0

(2x - 9)(x + 5) = 0

x = 4.5, -5.

So x = 4.5 ( as the width cannot be negative).

Therefore the dimenions of the rectangle are

width = 4.5 ft, length = 2(4.5) + 1 = 10 ft.

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Brainliest! :) The length of a rectangle is 1 ft more than twice the width, and the area of the rectangle is 45 ft^2. Find the dimensions of the rectangle.

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Brainliest! :)

The length of a rectangle is 1 ft more than twice the width, and the area of the rectangle is 45 ft^2. Find the dimensions of the rectangle.