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Home » Thus, the intersection point is \(\left(\frac23, \frac133\right)\). - Veritas Home Health

Thus, the intersection point is \(\left(\frac23, \frac133\right)\). - Veritas Home Health

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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r^2 \leq \frac{100}{1} = 100 \Rightarrow r \leq 10.
But we must check if this maximum is **attainable**. That is, does the...
Set $\sin(2\theta) = 1 \Rightarrow 2\theta = \frac{\pi}{2} + 2k\pi \Ri...
Try $\theta = \frac{\pi}{4}$:
Then $r^2 = \frac{100}{1} = 100 \Rightarrow r = 10$.
Check if this point lies within the circular region:
The circle has radius 10, so $r = 10$ is on the boundary — valid.
Thus, the **maximum possible value of $r$** is:
\boxed{10}
Question: A science policy analyst is modeling energy transmission eff...

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