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Answer step wise

Solution,
Given,
Radius of curvature, R = 3.0 m. Thus, focal length of the mirror, f = R/2 = +1.5 m 
Object distance, u = - 5.0 m
Let the image distance be v.
Using the mirror equation,
​​​​$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}+\frac{1}{-5}=\frac{1}{1.5} \\ \frac{1}{v}=\frac{2}{3}+\frac{1}{5}=\frac{10+3}{15}=\frac{13}{15} \\ v=\frac{15}{13}=1.15m \end{gathered}$$

The image is virtual, erect and diminished.