Answer:
a) in divergent lenses all images are virtual , b) The right image, c) q = 28 cm  d)   h ’= 4.4 cm
Explanation:
For this exercise we will use the equation the constructor
    1 / f = 1 / p + 1 / q
Where f is the focal length, p the distance to object and q the distance to the image
   1 / q = 1 / f - 1 / p
As the lens is divergent the focal length is negative
   1 / q = 1 / (- 20) - 1/70
   1 / q = -0.05 - 0.014286 = 0.0643
   q = -15.56 cm
For a graphic analysis see attached
a) in divergent lenses all images are virtual
b) The right image
c) q = 28 cm
d) let's use the magnification equation
   M = h ’/ h = - q / p
   h ’= -q / p h
   h ’= - (-15.56) / 70 20
   h ’= 4.4 cm
Being positive indicates that the image is right