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Home » American Pie: The Ultimate Guide to Every Epic Romance That Ruined Performances! - Abbey Badges

American Pie: The Ultimate Guide to Every Epic Romance That Ruined Performances! - Abbey Badges

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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🔥 Popular Posts

\(n = 10\), First term \(a = 3\), Common difference \(d = 5\)
\(S_{10} = rac{10}{2}(2 imes 3 + 9 imes 5)\)
\(S_{10} = 5(6 + 45) = 5 imes 51 = 255\)
Question:** Solve for \(x\) in the equation \(2x^2 - 8x + 6 = 0\) usin...
Quadratic formula: \(x = rac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(b^2 - 4ac = (-8)^2 - 4 imes 2 imes 6 = 64 - 48 = 16\)
\(x = rac{-(-8) \pm \sqrt{16}}{4}\)
\(x = rac{8 \pm 4}{4}\)
\(x = rac{12}{4} = 3\) or \(x = rac{4}{4} = 1\)
\(x = 3\) and \(x = 1\)

📝 Recent Posts

Thus, the answer is $\boxed{437}$.
Question: What is the smallest positive integer whose square ends in 2...
Solution: Let $x^2 \equiv 25 \pmod{100}$.
This implies $x \equiv \pm5 \pmod{10}$. Testing $x = 5$: $5^2 = 25$ en...
Thus, the smallest such integer is $\boxed{5}$.Question: An entomologi...
Solution: The problem requires calculating the number of distinct perm...
\frac{10!}{5! \cdot 3! \cdot 2!} = \frac{3628800}{120 \cdot 6 \cdot 2}...
The number of distinct observation sequences is $\boxed{2520}$.
Question: A science communicator creates a 10-part video series with 4...
Solution: This is a multinomial permutation problem. The total sequenc...

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