Final answer:
To sketch the conic r = 10/(3-2sin(theta)) and label the vertices, we can identify it as an ellipse with a shifted angle of 30 degrees. The vertices are (10/3, 0) and (-10/3, pi).
Step-by-step explanation:
To sketch the conic defined by the equation r = 10/(3-2sin(theta)), we can start by identifying the type of conic. In this case, the equation represents an ellipse. The formula for an ellipse in polar coordinates is r = a(1 - e^2)/(1 - ecos(theta)), where a is the semi-major axis and e is the eccentricity.
A term in the form 2sin(theta) appears in the given equation, which can be rewritten as sin(theta) = 1/2. This indicates the presence of a shift of pi/6 in the angle theta. Since sin(pi/6) = 1/2, the ellipse is shifted 30 degrees counter-clockwise.
The vertices of the ellipse can be found by plugging in theta = 0 and theta = pi into the equation. When theta = 0, r becomes 10/(3 - 2sin(0)) = 10/3. When theta = pi, r becomes 10/(3 - 2sin(pi)) = -10/3. So the two vertices of the ellipse are (10/3, 0) and (-10/3, pi).