Question

Sam is transferring files from his friend's laptop into his flash drive. This is the formula for the size of the files on Sam's friends drive S (measured in megabytes) as a function of time t (measured in seconds): S(t)=5t+45

Use the above information to answer the following:
When the transfer began the drive had___ megabytes of files on it

Every 10 seconds, an additional___ megabytes are transferred into the drive.

Answer 1

1. 45 megabytes

2. 50 megabytes

Explanation:

Given

S(t) = 5t + 45

Required

Find S(t) when the transfer began

And find the additional transfer size at interval of 10 seconds

When the file transfer initially began, the value of t is 0.

Given that S(t)=5t+45.

Substitute 0 for t

S(0)=5 * 0 + 45

S(0) = 0 + 45

S(0) = 45.

Hence, the hard drive had 45 megabytes on it when the transfer started.

At every interval of 10 seconds. We take t to be 10, 20, 30...seconds

When t = 10

S(t)=5t+45 becomes

S(10) = 5(10) + 45

S(10) = 50 + 45

S(10) = 95 megabytes

When t = 20

S(t)=5t+45 becomes

S(20) = 5(20) + 45

S(20) = 100 + 45

S(20) = 145 megabytes

When t = 30

S(t)=5t+45 becomes

S(30) = 5(30) + 45

S(30) = 150 + 45

S(30) = 195 megabytes

Calculating the difference

Difference = S(30) - S(20) or S(20) - S(10)

Difference = 195 - 145 or 145 - 95

Difference = 50 or 50

See that the difference is the same.

So, at every 10 seconds interval, there's an additional 50 megabytes transferred