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x2 + 13x

Q: The length of a rectangular backyard is 13 feet longer than its width x What is the area of the backyard in terms of x?

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x2+13x

piece a has a greater length than piece b (a>b)

Kilogram is weight, not volume.

Let the width of the rectangular prism be y Therefore we can write all the measurements of the rectangular prism in terms of y. width = y length = 2y height = y + 4 Finding the volume of the rectangular prism y x 2y x (y + 4) = 48 2y2(y+4) = 48 y2(y+4) = 24 From this equation, we can see that y must equal 2 (22 x 6 = 24) Therefore the width of the rectangular prism is 2m. Note that in this question, when we got to the equation y2(y+4) = 24, we used 'guess and check' to find that y was. We purposely avoided expanding the bracket as this would have left us with a cubic to solve, which is quite difficult to solve analytically unless you already know one of its roots.

This appears to be a comparison of two similar triangles. Measure the length of a corresponding side of each triangle. Let the side having the shorter length be b, and c the side having the longer length. Then the scale is b : c or b/c If possible multiply or divide the numbers forming the ratio to provide an answer in its lowest terms.

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x2+13x

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piece a has a greater length than piece b (a>b)

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x^2 + 2x ft

Kilogram is weight, not volume.

Let the width of the rectangular prism be y Therefore we can write all the measurements of the rectangular prism in terms of y. width = y length = 2y height = y + 4 Finding the volume of the rectangular prism y x 2y x (y + 4) = 48 2y2(y+4) = 48 y2(y+4) = 24 From this equation, we can see that y must equal 2 (22 x 6 = 24) Therefore the width of the rectangular prism is 2m. Note that in this question, when we got to the equation y2(y+4) = 24, we used 'guess and check' to find that y was. We purposely avoided expanding the bracket as this would have left us with a cubic to solve, which is quite difficult to solve analytically unless you already know one of its roots.

In mathematical terms there is no limit. In practical terms, construction techniques and the structural strength of materials will impose some limit.

Measure the length and breadth of one face of the rectangle. Measure the thickness of the rectangle hollow piece. Multiply the length, breadth and thickness and this will give you the volume of the rectangular hollow piece in terms of cubic units. That is, if you measured the length, breadth and height in centimeters, the volume will be in cubic centimeters. Example: If the length of the piece is 10 cm, the breadth of the piece is 6 cm, and the thickness of the piece is 5 cm, the volume of the rectangular hollow piece is given by: 10 cm X 6 cm X 5 cm = 300 cubic cm or cm3