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The length of a rectangle is 1 inch more than 2 times its width.The area of the rectangle is 36 square inches. what is the length of the rectangle?

Correct Answer: 9 inches
Let the width be W. Then, Length = 2W + 1 Now, area,  (2W + 1)×W = 36 or, 2W2  + W = 36 or, 2W2  + W - 36 = 0 or, 2W2  + 9W - 8W - 36 = 0 or, W(2W + 9) - 4(2W + 9) = 0 or, (2W + 9)(W - 4) = 0 or, W - 4 = 0 [2W + 9≠0] or, W = 4 So, length, 2W + 1 = 2×4 + 1 =  9 inches

Simplify the value of $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}} = ?$$

The size of the philips LCD television screen is given as the length of the screen's diagonal , Assume that the screens of the television were flat. Find out how many square inches greater would be the screen of square inches greater would be the screen of a square 21 -inch television set than that of a square 19-inch television set.

$$\frac{{38 \times 38 \times 38 + 34 \times 34 \times 34 + 28 \times 28 \times 28 - 38 \times 34 \times 84}}{{38 \times 38 + 34 \times 34 + 28 \times 28 - 38 \times 34 - 34 \times 28 - 38 \times 28}}$$            is equal to = ?

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