f = 15cm (diverging mirror); u = -35cm(because object is in front of the mirror) v=? ⇒\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\) ⇒ \(\frac{1}{15}\) = \(\frac{1}{v}\) - \(\frac{1}{35}\) ⇒ \(\frac{1}{15}\) + \(\frac{1}{35}\) = \(\frac{1}{v}\) ⇒ \(\frac{105}{10}\) = \(\frac{1}{v}\) v = \(\frac{105}{10}\) = 10.5cm ∴The image is 10.5cm behind the mirror, since v is positive.