5. as a radiology specialist, you use the radioactive substance \iodine - 131\ to diagnose conditions of the thyroid gland. the amount of radioactive substance decays with time. your hospital currently has a 20 - gram supply of iodine - 131. the following function models the number of grams remaining after x days. y = 20(0.917)^x
a. determine the daily percentage decay rate.
b. determine how much substance remains after 5 days. round to the nearest whole number of grams.
c. determine the half - life of iodine - 131 by finding how many days it takes for the 20 - gram supply to decrease to 10 grams. round to the nearest whole number of days. (you can use a graph, table, or guess & check.)
d. explain the practical meaning of the horizontal asymptote in this situation.
6. a child currently receives a weekly allowance of $3 and asks their parents for a raise. the parent offers the child two options:
option 1: the child gets a $0.25 raise each week
option 2: the child gets a 5% raise each week
a. write an equation for each option to determine the weekly allowance y after x weeks of raises.
b. use technology to graph your two functions together to compare them. which option is better at first? after how many weeks will the other option begin to offer a bigger allowance?

Question

Accepted Answer

### 5.
#### a.
# Explanation:
## Step1: Identify decay - factor formula
The general form of an exponential decay function is $y = a(1 - r)^x$, where $r$ is the decay rate. Given $y = 20(0.917)^x$, we set $1 - r=0.917$.
## Step2: Solve for $r$
$r = 1 - 0.917=0.083$. To convert to a percentage, we multiply by 100. So the daily percentage decay rate is $8.3\%$.
# Answer:
$8.3\%$

#### b.
# Explanation:
## Step1: Substitute $x = 5$ into the function
We have the function $y = 20(0.917)^x$. Substitute $x = 5$: $y=20(0.917)^5$.
## Step2: Calculate the value
$y = 20	imes0.917^5=20	imes0.66342 = 13.2684\approx13$ grams.
# Answer:
$13$ grams

#### c.
# Explanation:
## Step1: Set up the equation
We want to find $x$ when $y = 10$ in the equation $y = 20(0.917)^x$. So, $10 = 20(0.917)^x$.
## Step2: Simplify the equation
Divide both sides by 20: $\frac{10}{20}=(0.917)^x$, or $0.5=(0.917)^x$.
## Step3: Take the natural - logarithm of both sides
$\ln(0.5)=x\ln(0.917)$.
## Step4: Solve for $x$
$x=\frac{\ln(0.5)}{\ln(0.917)}=\frac{- 0.693147}{-0.08667}=8.009\approx8$ days.
# Answer:
$8$ days

#### d.
# Explanation:
The exponential decay function is $y = 20(0.917)^x$. As $x
ightarrow\infty$, $y
ightarrow0$. The horizontal asymptote is $y = 0$. In a practical sense, this means that as time ($x$, in days) goes on indefinitely, the amount of the radioactive iodine - 131 will approach zero grams. Eventually, the radioactive substance will completely decay.
# Answer:
As time goes on indefinitely, the amount of radioactive iodine - 131 will approach zero grams.

### 6.
#### a.
# Explanation:
## Option 1:
For a linear increase, the initial amount $a = 3$ and the common difference $d=0.25$. The linear equation is $y=3 + 0.25x$.
## Option 2:
For a percentage increase, the initial amount $a = 3$ and the growth factor $r = 1+0.05 = 1.05$. The exponential equation is $y = 3(1.05)^x$.
# Answer:
Option 1: $y = 3+0.25x$; Option 2: $y = 3(1.05)^x$

#### b.
# Explanation:
## Step1: Analyze the initial situation
For $x = 1$:
- Option 1: $y=3 + 0.25	imes1=3.25$.
- Option 2: $y=3	imes1.05 = 3.15$. So Option 1 is better at first.
## Step2: Find when Option 2 overtakes Option 1
We can use a graphing utility or solve the equation $3+0.25x=3(1.05)^x$. By trial - and - error or using a graphing calculator:
When $x = 10$:
- Option 1: $y=3+0.25	imes10=3 + 2.5 = 5.5$.
- Option 2: $y=3	imes(1.05)^{10}=3	imes1.628894626777=4.88668388033$.
When $x = 11$:
- Option 1: $y=3+0.25	imes11=3 + 2.75 = 5.75$.
- Option 2: $y=3	imes(1.05)^{11}=3	imes1.710339358116=5.13101807435$.
When $x = 12$:
- Option 1: $y=3+0.25	imes12=3 + 3=6$.
- Option 2: $y=3	imes(1.05)^{12}=3	imes1.795856326022=5.38756897807$.
When $x = 13$:
- Option 1: $y=3+0.25	imes13=3 + 3.25 = 6.25$.
- Option 2: $y=3	imes(1.05)^{13}=3	imes1.885649142323=5.65694742697$.
When $x = 14$:
- Option 1: $y=3+0.25	imes14=3 + 3.5 = 6.5$.
- Option 2: $y=3	imes(1.05)^{14}=3	imes1.979931599439=5.93979479832$.
When $x = 15$:
- Option 1: $y=3+0.25	imes15=3 + 3.75 = 6.75$.
- Option 2: $y=3	imes(1.05)^{15}=3	imes2.078928179411=6.23678453823$.
When $x = 16$:
- Option 1: $y=3+0.25	imes16=3 + 4=7$.
- Option 2: $y=3	imes(1.05)^{16}=3	imes2.182874588382=6.54862376514$.
When $x = 17$:
- Option 1: $y=3+0.25	imes17=3 + 4.25 = 7.25$.
- Option 2: $y=3	imes(1.05)^{17}=3	imes2.292018317801=6.8760549534$.
When $x = 18$:
- Option 1: $y=3+0.25	imes18=3 + 4.5 = 7.5$.
- Option 2: $y=3	imes(1.05)^{18}=3	imes2.406619233691=7.21985770107$.
So after 18 weeks, Option 2 will begin to offer a bigger allowance.
# Answer:
Option 1 is better at first. After 18 weeks, Option 2 will begin to offer a bigger allowance.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

5.

a.

Step1: Identify decay - factor formula

Step2: Solve for $r$

Step1: Substitute $x = 5$ into the function

Step2: Calculate the value

Step1: Set up the equation

Step2: Simplify the equation

Step3: Take the natural - logarithm of both sides

Step4: Solve for $x$

Answer:

b.