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Home » Daher ist der kleinste Primfaktor von 135 \(\boxed3\). - Get link 4share

Daher ist der kleinste Primfaktor von 135 \(\boxed3\). - Get link 4share

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 $2c + 4d = 8$.

📌 Solving equations 1 and 2: From 1, $a = 5 - 3b$. Substitute into 2: $2(5 - 3b) + 4b = 6 \Rightarrow 10 - 6b + 4b = 6 \Rightarrow -2b = -4 \Rightarrow b = 2$, then $a = 5 - 6 = -1$.

📌 Solving equations 3 and 4: From 3, $c = 7 - 3d$. Substitute into 4: $2(7 - 3d) + 4d = 8 \Rightarrow 14 - 6d + 4d = 8 \Rightarrow -2d = -6 \Rightarrow d = 3$, then $c = 7 - 9 = -2$.

📌 Thus, $\mathbf{M} = \begin{pmatrix} -1 & 2 \\ -2 & 3 \end{pmatrix}$. Final answer: $\boxed{\begin{pmatrix} -1 & 2 \\ -2 & 3 \end{pmatrix}}$.

📌 Question: Find all angles $z \in [0^\circ, 360^\circ]$ that satisfy $2\sin(2z) = \sqrt{3}$.

📌 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: $\boxed{2\sqrt{61}}$.

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