Agent Hunt is transferring classified files from the CIA mainframe into his flash drive. $\newline$ $S$ represents the size of the files on the drive (in megabytes) after $t$ seconds. $\newline$ $S=5 t+45$ $\newline$ How many megabytes are transferred every $10$ seconds? $\newline$ megabytes

View step-by-step help

Home

Math Problems

Algebra 2

Exponential growth and decay: word problems

Full solution

Q. Agent Hunt is transferring classified files from the CIA mainframe into his flash drive.

\newline

S

represents the size of the files on the drive (in megabytes) after

t

seconds.

\newline

S=5 t+45

\newline

How many megabytes are transferred every

10

seconds?

\newline

megabytes

Identify formula and question: Identify the given formula and what is asked. $\newline$ The formula given is $S = 5t + 45$ , where $S$ represents the size of the files on the drive (in megabytes) after $t$ seconds. We need to find out how many megabytes are transferred every $10$ seconds.
Determine size at start of interval: Determine the size of the files on the drive at the start of the $10$ -second interval. $\newline$ To find out how many megabytes are transferred every $10$ seconds, we need to calculate the size of the files at the beginning and at the end of a $10$ -second interval. Let's start with the size at the beginning, which is when $t = 0$ seconds. $\newline$ $S = 5(0) + 45$ $\newline$ $S = 0 + 45$ $\newline$ $S = 45$ megabytes
Determine size at end of interval: Determine the size of the files on the drive at the end of the $10$ -second interval. $\newline$ Now, we calculate the size of the files at the end of the $10$ -second interval, which is when $t = 10$ seconds. $\newline$ $S = 5(10) + 45$ $\newline$ $S = 50 + 45$ $\newline$ $S = 95$ megabytes
Calculate difference in size: Calculate the difference in size to find out how many megabytes are transferred. $\newline$ The number of megabytes transferred in $10$ seconds is the difference in size between the end and the beginning of the $10$ -second interval. $\newline$ Megabytes transferred = Size at end - Size at beginning $\newline$ Megabytes transferred = $95 \text{ megabytes} - 45 \text{ megabytes}$ $\newline$ Megabytes transferred = $50 \text{ megabytes}$