Solving for image height given ##h_o, d_i##, and r

[harrypoterfan]

Homework Statement

A convex mirror with a radius of curvature of −30.0 cm is used to form an image of an arrow that is 10.6 cm away from the mirror. If the arrow is 1.70 cm tall and inverted (pointing below the optical axis), what is the height of the arrow's image? (Include the sign of the value in your answer.)

Relevant Equations

$$f = -\frac{1}{r}$$
$$f^-1 = d_i^{-1} + d_0^{{-1}$$
$$m = \frac{h_i}{h_0} =-\frac{d_i}/{d_o}$$

$$f = -\frac{-30}{2} = 15$$
solving for $d_0$ in $$f^{-1} = d_i^{-1} + d_0^{-1}$$,
$$d_0 = (f^{-1} - d_i^{-1})^{-1}$$
= -36.1364
solving for $h_i$ in $$m = \frac{h_i}{h_0} =-\frac{d_i}{d_o}$$,
$$h_i = -d_i\times\frac{h_0}{d_o} = 0.4987$$
I'm told by webassign that it should be negative and that -0.4987 is off by more than 10%

 

[TSny]

Be sure you are following the sign conventions that are used in your course for mirrors. Usually, the conventions are such that the focal length of a convex mirror is negative.

You solved for $d_0$. But $d_0$ is given. You need to find $d_i$.