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about 170 ft

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Q: What is the width of the freedom tower?
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The length of a rectangle is 3 inches more than its width and its perimeter is 22 inches Find the width of the rectangle?

Perimeter = length + width + length + width = 2 x (length + width) Given: perimeter = 22in length = width + 3in Thus 22 = 2 x (width + 3 + width) 11 = 2 x width + 3 8 = 2 x width 4 = width So the width is 4in.


What are the dimensions of a rectangle with length 5 in longer than its width which has 1 inside squares cut from the corners and the flaps folded up to form an open box of volume 500 cu in?

It is 22 in by 27 in. ------------------------------------------- How it is solved: The rectangle is 5 in longer than its width, so the dimensions can be given in terms of its width: Its dimensions are width by (width+5) In removing 1 inch squares from the corners, he flaps are going to be (width-2) and (width+5-2) = (width-3) long. Thus the box formed will have volume: volume = (width - 2) in × (width + 3) in × 1 in → volume = width² + width - 6 in³ But we are told this is 500 in³; thus: width² + width - 6 in³ = 500 in³ → width ² + width - 506 = 0 → (width + 23)(width - 22) = 0 → width = -23 or 22 As a length cannot be negative, the width must be 22 in which means the original rectangle is 22 in by 22+5 in = 27 in


If a rectangle has an area of 24m² and it length is 6m What is its width?

Area = Length * Width 24 = 6 * Width so width = 4m


What is the area of a rectangle whose diagonal is 17 cm with a perimeter of 460 mm explaining work with answer?

The area of rectangle is : 7820.0


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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