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Home » Thus, the dimensions are width \(\frac143\) m and length \(\frac373\) m. - Network4

Thus, the dimensions are width \(\frac143\) m and length \(\frac373\) m. - Network4

📅 May 19, 2026 👤 scraface

May 19, 2026

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📝 Recent Posts

The sum of the first \(n\) positive integers is given by the formula \...
\(S = \frac{50(50+1)}{2} = \frac{50 \times 51}{2} = \frac{2550}{2} = 1...
A rectangular garden has a length that is 3 times its width. If the pe...
Perimeter: \(2(w + 3w) = 64\)
Solve the quadratic equation \(2x^2 - 4x - 6 = 0\) using the quadratic...
Quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Here, \(a = 2\), \(b = -4\), \(c = -6\)
Discriminant: \(b^2 - 4ac = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48...
\(x = \frac{4 \pm \sqrt{64}}{4} = \frac{4 \pm 8}{4}\)
Solutions: \(x = \frac{12}{4} = 3\) and \(x = \frac{-4}{4} = -1\)

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