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Area 36x4-64x2 and width is 6x2-8x Area = Length * width
36x4 -64x2 = L * 6x2 -8x

(36x4 -64x2) / (6x2 -8x) = L

(36x4 -64x2) / (6x2 -8x) = L

((6x2 -8x)2 + 96x3)/ (6x2 -8x) = L

1 + 96x3/ (6x2 -8x) = L
1 + 96x3/ x(6x -8) = L
1 + 96x2/ (6x -8) = L

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What is the length and width of a rectangle that has a perimeter of 20 inches and an area of 24.4524 square inches showing work with final answers?

What do we know about the perimeter of a rectangle? perimeter = 2 × (length + width) → 2 × (length + width) = 20 in → length + width = 10 in → length = 10 in - width What do we know about the area of a rectangle: area = length × width → length × width = 24.4524 in² But from the perimeter we know the length in terms of the width and can substitute it in: → (10 in - width) × width = 24.4524 in² → 10 in × width - width² = 24.4524 in² → width² - 10 in × width + 24.4524 in² = 0 This is a quadratic which can be solved by using the formula: ax² + bx + c → x = (-b ±√(b² - 4ac)) / (2a) → width = (-10 ±√(10² - 4 × 1 × 24.4524)) / (2 × 1) in → width = -5 ± ½√(100 - 97.8096) in → width = -5 ±½√2.1904 in → width = -5 ± 0.74 in → width = 4.26 in or 5.74 in → length = 10 in - 4.26 in = 5.74 in or 10 in - 5.74 in = 4.26 in (respectively) By convention the width is the shorter length (though it doesn't have to be) making the width 4.26 in and the length 5.74 in. Thus the rectangle is 5.74 in by 4.26 in


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What is the width and length of a rectangle if the perimeter is 50 and the length is 5 cm more than the width?

Perimeter = 2 lengths and 2 widths In your case length + width = 25cm If length is 5cm more than width then length = 15cm and width = 10cm

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What is the length and width of a rectangle that has a perimeter of 20 inches and an area of 24.4524 square inches showing work with final answers?

What do we know about the perimeter of a rectangle? perimeter = 2 × (length + width) → 2 × (length + width) = 20 in → length + width = 10 in → length = 10 in - width What do we know about the area of a rectangle: area = length × width → length × width = 24.4524 in² But from the perimeter we know the length in terms of the width and can substitute it in: → (10 in - width) × width = 24.4524 in² → 10 in × width - width² = 24.4524 in² → width² - 10 in × width + 24.4524 in² = 0 This is a quadratic which can be solved by using the formula: ax² + bx + c → x = (-b ±√(b² - 4ac)) / (2a) → width = (-10 ±√(10² - 4 × 1 × 24.4524)) / (2 × 1) in → width = -5 ± ½√(100 - 97.8096) in → width = -5 ±½√2.1904 in → width = -5 ± 0.74 in → width = 4.26 in or 5.74 in → length = 10 in - 4.26 in = 5.74 in or 10 in - 5.74 in = 4.26 in (respectively) By convention the width is the shorter length (though it doesn't have to be) making the width 4.26 in and the length 5.74 in. Thus the rectangle is 5.74 in by 4.26 in


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What are the dimensions of a table if the perimeter is 36 and the length is 5 times the width?

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