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Home » Solution: The cross product \(\mathbfv \times \mathbfa\) is computed as: - Abu Waleed Tea

Solution: The cross product \(\mathbfv \times \mathbfa\) is computed as: - Abu Waleed Tea

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 \( a_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2 \)

📌 The 10th term is \( a_{10} = 6(10) + 2 = 60 + 2 = 62 \).

📌 A circle is inscribed in a right triangle with legs of 8 and 15. Find the radius of the circle.

📌 The hypotenuse \( c = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \).

📌 The area \( A \) of the triangle is \( rac{1}{2} imes 8 imes 15 = 60 \).

📌 The semi-perimeter \( s = rac{8 + 15 + 17}{2} = 20 \).

📌 The radius \( r \) of the inscribed circle is given by \( r = rac{A}{s} = rac{60}{20} = 3 \).

📌 Solve for \( x \) in the equation \( \log_2(x+1) + \log_2(x-1) = 3 \).

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