(4x² + 6) + (2x² + 6x + 3) = 4x² + 6 + 2x² + 6x + 3 = 6x² + 6x + 9

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Answer: 6x² + 6x + 9 (Answer C)
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7 0

3 years ago

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The width of a rectangular parking lot is 53 ft less than its length. Determine the dimensions of the parking lot if it measures

Sever21 [200]

Answer:

The dimension of the lot is 201.28 ft by 148.28 ft

Step-by-step explanation:

Given;

diagonal of the parking lot, d = 250 ft

let the length of the parking lot = L

the width, W = L - 53

The diagonal of the lot, length of the lot, and width of the lot form a right triangle.

Apply Pythagoras theorem to determine the length, L

L² + W² = 250²

L² + (L - 53)² = 250²

L² + L² - 106L + 2809 = 62,500

2L² - 106L + 2809 - 62,500 = 0

2L² - 106L - 59,691 = 0

this forms quadratic equation; a = 2, b = -106 and c = -59,691

$L = \frac{- b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\L = \frac{- (-106) \ \ +/- \ \ \sqrt{(-106)^2 - 4(-59691 \times 2)} }{2(2)}\\\\L = \frac{106 \ \ +/- \ \ \sqrt{488764} }{4} \\\\L = \frac{106 + 699.12}{4} \\\\L = 201.28 \ ft$

The width, W = 201.28 - 53

W = 148.28 ft

7 0

2 years ago