Sophia has an ear infection. The doctor prescribes a course of antibiotics. Sophia is told to take 500 mg doses of the antibiotic regularly every 12 hours for 10 days. Sophia is curious and wants to know how much of the drug will be in her body over the course of the 10 days. She does some research online and finds out that at the end of 12 hours, about 4.5% of the drug is still in the body. What quantity of the drug is in the body right after the first dose, the second dose, the third dose, the fourth dose (10 points)? When will the total amount of the antibiotic in Sophia’s body be the highest? What is that amount (10 points)? Answer Sophia’s original question: Describe how much of the drug will be in her body at various points over the course of the 10 days (10 points).

As Sophie has just taken the medicine, the quantity of drug present in her body is 500mg

Amount left from 1st dose = 4.5/100 × 500 = 22.5mg

Amount of Drug after 2nd dose = Amount left from 1st dose + 2nd dose taken.

Amount of Drug after 2nd dose = 22.5mg + 500mg

Amount of Drug after 2nd dose = 522.5mg

Amount left from 2nd dose = 4.5/100 × 522.5 = 23.513 mg

Amount of Drug after 3rd dose = Amount left from 2nd dose + 3rd dose taken.

Amount of Drug after 3rd dose = 23.513mg + 500mg

Amount of Drug after 3rd dose = 523.513 mg

Amount left from 3rd dose = 4.5/100 × 523.513 = 23.558mg

Amount of Drug after 4rth dose = Amount left from 3rd dose + 4rth dose taken.

Amount of Drug after 4rth dose = 23.558mg + 500mg

Amount of Drug after 4rth dose = 523.558 mg

The pattern formed above can be written in the form of sum of finite geometric series:

Sₙ = 500 + 500(0.045) +500(0.045)² + ........................500(0.045)ⁿ

Sₙ = 500(1 + 0.045 + 0.045² ........................0.045ⁿ)

Where

Sₙ = Sum till nth term

a = 500

r = 0.045

As Sophia takes medicines 2 times a day, she takes medicine 20 times in 10 days. So substitute n=20 in the formula for the sum of finite geometric series

Greatest amount of antibiotic present = 523.5602 mg

This amount is almost constant after 3rd dose, so it will occur after every dose after 3rd dose

After 1st dose = 500mg

After 2nd dose = 522,5mg

After 3rd dose = 523.513 mg

After the the amount left in the body almost remains constant.

Which means that after each dose, the amount in body is maximum, at the end of the 12 hours, the amount reduces to 4.5%, which reaches almost the same maximum after taking another dose.

let's solve your equation step-by-step.

4x+2=2x+12

step 1: subtract 2x from both sides.

4x+2−2x=2x+12−2x

2x+2=12

step 2: subtract 2 from both sides.

2x+2−2=12−2

2x=10

step 3: divide both sides by 2.

2x

2

=

10

2

x=5

step-by-step explanation:

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .