the amount (in milligrams) of a drug in the body t hours after taking a pill is given by A(t)=30(0.8)t .

(a) What is the initial dose given?

(b) What percent of the drug leaves the body each hour?
Percent:

(c) What is the amount of drug left after 9 hours?

(d) What is the first whole number of hours at which there is less than 5 milligrams left in the body?

(a) The initial dose given is 30 milligrams.

(b) Each hour, approximately 20% of the drug leaves the body. So, the percent of the drug that leaves the body each hour is 20%.

(c) To find the amount of drug left after 9 hours, we substitute t = 9 into the equation A(t) = 30(0.8)^t:
A(9) = 30(0.8)^9 ≈ 3.874 milligrams

(d) To find the first whole number of hours at which there is less than 5 milligrams left in the body, we need to find the value of t that satisfies A(t) < 5. Let's begin checking values:
A(1) = 30(0.8)^1 = 24 milligrams
A(2) = 30(0.8)^2 = 19.2 milligrams
A(3) = 30(0.8)^3 = 15.36 milligrams

At t = 3, we have less than 5 milligrams left in the body. So, the first whole number of hours at which there is less than 5 milligrams left in the body is 3 hours.

(a) The initial dose given is the amount of the drug in the body at t=0.

Substituting t=0 into the equation A(t)=30(0.8)t:
A(0) = 30(0.8)^0 = 30(1) = 30 milligrams.

(b) The percent of the drug that leaves the body each hour can be calculated by finding the ratio of the decrease in the amount of the drug in one hour to the initial dose.
The decrease in the amount of the drug in one hour is 30(0.8) because each hour, the amount decreases by 20% (0.8) of the previous amount.
So, the percent is (30(0.8)/30)*100 = (24/30)*100 = 80%.

(c) To find the amount of drug left after 9 hours, we substitute t=9 into the equation A(t)=30(0.8)t:
A(9) = 30(0.8)^9 ≈ 6.663 milligrams (rounded to three decimal places).

(d) To find the first whole number of hours at which there is less than 5 milligrams left in the body, we need to solve the equation A(t) < 5.
Substituting A(t)=5 into the equation A(t)=30(0.8)t:
5 = 30(0.8)t
Dividing both sides by 30:
0.1667 = (0.8)t
Taking the logarithm with base 0.8 of both sides:
log0.8(0.1667) ≈ t
t ≈ 13.91.

Therefore, the first whole number of hours at which there is less than 5 milligrams left in the body is 14 hours.

To find the initial dose given, we need to evaluate the function A(t) at t=0.

(a) A(0) = 30(0.8)^0 = 30 * 1 = 30

Therefore, the initial dose given is 30 milligrams.

To find the percent of the drug that leaves the body each hour, we need to determine the decay factor, which is represented by 0.8 in the function A(t)=30(0.8)^t.

(b) The decay factor can be thought of as the percentage remaining. So, to find the percentage leaving the body each hour, we subtract the decay factor from 1 and multiply by 100.

Percent leaving the body each hour = (1 - 0.8) * 100 = 0.2 * 100 = 20%

Therefore, 20% of the drug leaves the body each hour.

To find the amount of drug left after 9 hours, we need to evaluate the function A(t) at t = 9.

(c) A(9) = 30(0.8)^9 ≈ 30 * 0.134217728 = 4.02653284

Therefore, the amount of drug left after 9 hours is approximately 4.03 milligrams.

To find the first whole number of hours at which there is less than 5 milligrams left in the body, we can set up an inequality using the function A(t).

(d) A(t) < 5

30(0.8)^t < 5

Now, we can solve this inequality for t.

Taking the natural logarithm of both sides, we get:

ln(30) + tln(0.8) < ln(5)

Dividing both sides by ln(0.8), we have:

t < (ln(5) - ln(30)) / ln(0.8)

Using a calculator, we can evaluate the right side of the inequality to find the first whole number that satisfies the inequality.

The first whole number of hours at which there is less than 5 milligrams left in the body is t = 12.

Therefore, the answer is 12 hours.

(a) at t=0, .8^0 = 1

(b) .8 means decrease of 20%

(c) 30*.8^9 = ?

(d) 30*.8^t = 5
.8^t = 1/6
t log(.8) = log(1/6)
t = log(1/6)/log(.8) = 8.03