An object is located at 51 millimeters from a diverging lens the object has a height of 13 millimeters and image height is 3.5 millimeters how far in front of the lens is the image located

Magnification by height of image:      

Image height/object height = 3.5/13

Magnification = v/u             v = image distance, u = object distance.

(3.5/13) = v/51

v/51 = (3.5/13)

v = 51*(3.5/13)

v ≈ 13.73 mm  

Because shotguns bullets ricochet

The best answer i can think of is b