Question 19 • Precalculus
For r = 2 + 5 sin θ, at θ = 0 we get r = 2, giving the point (2, 0). At θ = π/2, r = 7, and at θ = 3π/2, r = −3; a negative radius of −3 at 3π/2 is equivalent to (3, π/2). Choice D includes these points. Its point (−2, π) is also equivalent to (2, 0), so it represents a point on the curve.
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