An object 3 cm tall is placed 28 cm in front of a concave mirror of focal length 21 cm…

Step 1. The image. $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{21\ \text{cm}} + \dfrac{1}{28\ \text{cm}} = \dfrac{-4 + 3}{84\ \text{cm}} = -\dfrac{1}{84\ \text{cm}}$, so $v = -84\ \text{cm}$.

Step 2. The magnification. $m = -\dfrac{v}{u} = -\dfrac{-84\ \text{cm}}{-28\ \text{cm}} = -3$.

Step 3. The height. $h' = m \times h = -3 \times 3\ \text{cm} = -9\ \text{cm}$.