Question

Faye has a drink with 120 milligrams of caffeine. Each hour, the amount of caffeine in her body will be 80% of the amount from the previous hour.

Use the drop-down menus to complete an inequality that can be solved to find how much time, t, it will take for there to be less than 10 milligrams of caffeine remaining in Faye's body.

Answer 1

Answer: The other answer is right but for a quick answer it is in the image attached

Explanation:

Answer 2

12 hours are required to be less than 10 milligrams of caffeine in Faye's body

Explanation:

Geometric sequences

We can recognize a geometric progression, when we can get each member n as the previous member n-1 multiplied or divided by a constant value, called the common ratio.

Faye holds 120 milligrams of caffeine. We know each hour the amount of caffeine in her body will be 80% of the amount from the previous hour. For example, the first hour she will hold 80%(120)=96 milligrams of caffeine. Next hour it will be 80%(96)=76.8 and so on

We can see the sequence of contents of caffeine follows the rule of a geometric sequence and the common ratio is 80%, i.e. 0.8. The first term is 120, so the general term of the sequence is

`a_t=120(0.8)^t\ ,\ t>=0`

We are required to find how much time is needed to be less than 10 milligrams of caffeine remaining in Faye's body. It can be stated that

`120(0.8)^t<10`

Let's solve for t. Operating

`(0.8)^t<(10)/(120)`

Taking logarithms in both sides

`ln(0.8)^t<ln(10)/(120)`

Applying the power property of logarithms

`t\ ln(0.8)<ln(10)/(120)`

Solving for t. Since ln0.8 is negative, the direction of the inequality changes

`t>(ln(10)/(120))/(ln(0.8))`

`t>11.14`

Rounding to the next integer

12 hours are required to be less than 10 milligrams of caffeine in Faye's body