00 \, \text{cm}$ and the focal length of the diverging lens $f_2$ is $-20.0 \, \text{cm}$. The object is placed $2.50 \, \text{m}$ to the left of this combination. We can use the lens formula $1/f = 1/v - 1/u$ where $f$ is the focal length, $v$ is the image
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