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The fastest way to solve this problem is to know that a convex mirror and concave lens will have virtual, upright and reduced.

I just wanted to confirm this with calculations.
1/f=1/o+1/i
(1/-30)=(1/12)+(1/i) (focal length is -, because of convex mirror)
i=-8 ish
m=-i/o=8/12=0.66 <---- m=+ indicates upright=virtual, Since m is <1 it is reduced.

Is this the correct way of calculating this or do I put a negative in front of the do?

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