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 2/5 seconds?

Answer 1

Answer:

105 frames

Step-by-step explanation:

Given that 25 frames are played per second

25 frames= 1 sec

The video already played 3 and 2/5 seconds before the player started to count 0

Write 3 and 2/5 seconds as an improper fraction

=(5*3)+2 / 5 = 17/5 seconds

Multiply by 25 frames

17/5 *25 =85 frames

So according to the video counter,after 17/5 seconds,it should count 85 frames.However,at 0 seconds,it indicated a count of 190 frames.Thus,to get the number of frames that were already in count you subtract 85 frames from the 190 frames.

190-85=105 frames.

Answer 2

Answer:

The number of frames when the time was equal to $-3\frac{2}{5}$ seconds is 125.

Step-by-step explanation:

It is given that the player began counting at 0 seconds, even though the video had already played 190 frames. The video plays 25 frames per second.

The slope intercept form of linear function is

$y=mx+b$

where, m is rate of change and b is initial value.

It means the initial number of frames = 190

Change in frames per second = 25

The equation that represents the number of frames after x seconds is

$y=25x+190$            .... (1)

We need to find the number of frames when the time was equal to $-3\frac{2}{5}$ seconds.

$x=-3\frac{2}{5}=-\frac{13}{5}$

Substitute $x=-\frac{13}{5}$ in equation (1),

$y=25(-\frac{13}{5})+190$

$y=-65+190$

$y=125$

Therefore the number of frames when the time was equal to $-3\frac{2}{5}$ seconds is 125.