Using the magnification formula for mirrors: \[ m = \frac{f}{u-f} \] For the concave mirror, the object distance is \( u = -18 \, \text{cm} \), and the focal length is \( f = \frac{R}{2} = 6 \, \text{cm} \), where \( R = 12 \, \text{cm} \): \[ m_1 = \frac{6}{18 - 6} = \frac{1}{2} \] For the convex mirror, the object distance is the same, and the focal length is positive: \[ m_2 = \frac{6}{18 + 6} = \frac{1}{4} \] Hence, the ratio of the sizes of the images formed by the convex mirror and the concave mirror is: \[ \frac{m_2}{m_1} = \frac{1/4}{1/2} = \frac{1}{2} \] Thus, the correct answer is: \[ \frac{1}{2} \]
The net current flowing in the given circuit is ___ A.
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .