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Home » From Teasing Your Belly Button to Fowing: The Ultimate Guide to Navel Piercing Trends - ECD Germany

From Teasing Your Belly Button to Fowing: The Ultimate Guide to Navel Piercing Trends - ECD Germany

📅 May 15, 2026 👤 scraface

May 15, 2026

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📌 Solution: Divide both sides by 2: $\sin(2z) = \frac{\sqrt{3}}{2}$. The general solutions for $2z$ are $2z = 60^\circ + 360^\circ k$ or $2z = 120^\circ + 360^\circ k$, where $k$ is an integer. Solve for $z$: $z = 30^\circ + 180^\circ k$ or $z = 60^\circ + 180^\circ k$. Within $[0^\circ, 360^\circ]$, valid solutions are $z = 30^\circ, 60^\circ, 210^\circ, 240^\circ$. Final answer: $\boxed{30^\circ, 60^\circ, 210^\circ, 240^\circ}$.

📌 Question: In a city grid, $\|\overrightarrow{OA}\| = 5$ km and $\|\overrightarrow{OB}\| = 12$ km, with an angle of $90^\circ$ between them. If $\overrightarrow{OC} = 2\overrightarrow{OA} - \overrightarrow{OB}$, find $\|\overrightarrow{OC}\|$.

📌 Solution: Use the Pythagorean theorem since $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are perpendicular. Compute $\|\overrightarrow{OC}\|^2 = (2\|\overrightarrow{OA}\|)^2 + (\|\overrightarrow{OB}\|)^2 = 4(25) + 144 = 100 + 144 = 244$. Thus, $\|\overrightarrow{OC}\| = \sqrt{244} = 2\sqrt{61}$. Final answer: $\boxed{2\sqrt{61}}$.

📌 Question: There exist constants $m$ and $n$ such that $\overrightarrow{OC} = m\overrightarrow{OA} + n\overrightarrow{OB}$, where $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$, and $\overrightarrow{OC} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$. Enter the ordered pair $(m, n)$.

📌 Solution: Set up the system:

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