A convex mirror of radius of curvature 30cm forms an image that is...

Given $R = 30cm$

So, $f = \dfrac{R}{2} = + 15cm$

For convex mirror, virtual image is formed for real object. Thus, magnification $\left(m\right) = \dfrac{1}{2}$

The magnification can be written as

$m = - \dfrac{v}{u} . . . \left(1\right)$

Equation (1) implies that

$\dfrac{1}{2} = - \dfrac{v}{u} ⇒ v = - \dfrac{u}{2}$

The relation between the object distance, image distance and the focal length of the mirror is given by

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} . . . \left(2\right)$

Equation (2) implies that

$\dfrac{1}{15} = \dfrac{1}{- \dfrac{u}{2}} + \dfrac{1}{u} = - \dfrac{1}{u} ⇒ u = - 15cm$