Question

because of a problem in the program, the timer in a video player did not begin counting until the video had been playing for several seconds. the player began counting at 0 seconds, even though the video had already played 190 frames. the video plays 25 frames per second. how many frames had the video already played when the time was equal to $-3\frac{2}{5}$ seconds? \boxed{} frames

Explanation:

Step1: Convert mixed number to improper fraction

First, convert \(-3\frac{2}{5}\) to an improper fraction. \( -3\frac{2}{5}=-\frac{3\times5 + 2}{5}=-\frac{17}{5}\) seconds.

Step2: Calculate frames for negative time

The video plays 25 frames per second. So in \(\frac{17}{5}\) seconds (ignoring the negative sign for frame count calculation as we are looking at the time before the timer started), the number of frames is \(25\times\frac{17}{5}\).

Simplify \(25\times\frac{17}{5}\): \(25\div5 = 5\), then \(5\times17=85\) frames.

Step3: Subtract from initial frames

The timer started at 0 seconds when 190 frames had already played. For the time \(-3\frac{2}{5}\) seconds (which is before the timer started), we subtract the frames played in that negative time from 190. So \(190 - 85 = 105\) frames.

Answer:

105