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ramada plaza by wyndham albuquerque midtown - Network4

📅 May 18, 2026 👤 scraface

May 18, 2026

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📝 Recent Posts

The dot product is always zero, regardless of \(x\). Hence, the vector...
Thus, the solution is all real numbers \(x\), and the answer can be ex...
Find the sum of all angles \(z \in [0^\circ, 360^\circ]\) that satisfy...
We solve \(\sin z + \cos z = 1\) for \(z \in [0^\circ, 360^\circ]\).
Square both sides:
+ 2\sin z \cos z = 1 \Rightarrow 2\sin z \cos z = 0 \Rightarrow \sin 2...
So, \(2z = 0^\circ, 180^\circ, 360^\circ, 540^\circ, \ldots\), giving:
z = 0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ
Now check which satisfy the original equation \(\sin z + \cos z = 1\):
\(z = 0^\circ\): \(\sin 0 + \cos 0 = 0 + 1 = 1\) ✓

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