Question

question 3
1 pts
a standard equation applies as follows to find image distance (i) involving object distance (o) and focal length (f).
\\(\frac{1}{i}=\frac{1}{f}-\frac{1}{o}\\)
use the following values and calculations to determine what the appropriate numerator would be for the fraction after completing all operations.
\\(\frac{1}{i}=\frac{1}{28}-\frac{1}{308}=\frac{11}{308}-\frac{1}{308}=\frac{}{154}\\)
enter the appropriate value for the numerator

Explanation:

Step1: Analyze the fraction

We have the fraction $\frac{10}{154}$ (derived from $\frac{11}{308}-\frac{1}{308}=\frac{11 - 1}{308}=\frac{10}{308}=\frac{10\div2}{308\div2}=\frac{5}{154}$). The numerator is the number on the top - part of the fraction.

Answer:

5