Worksheet on exponential growth, decay, and compound interest with real-world application problems.

It looks like you've uploaded a worksheet with several problems related to exponential growth, decay, and interest models. Since I can't view images directly, I'll solve each problem based on the text you provided. Let's go through them one by one.

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Problem 1:

"Chadley invested $800 into an account that earned a fixed interest rate of 4.5% compounded monthly. How much will be in the account after 3 years?"

#### Solution:
We use the compound interest formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
Where:
- $ A $ = amount after time $ t $
- $ P = 800 $ (principal)
- $ r = 0.045 $ (annual interest rate)
- $ n = 12 $ (compounded monthly)
- $ t = 3 $ years

Plug in:
$$
A = 800 \left(1 + \frac{0.045}{12}\right)^{12 \times 3} = 800 \left(1 + 0.00375\right)^{36}
$$
$$
A = 800 (1.00375)^{36}
$$

Now calculate:
$$
(1.00375)^{36} \approx 1.1409
$$
$$
A \approx 800 \times 1.1409 = 912.72
$$

✔ Answer: Approximately $912.72

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Problem 2:

"Jake bought a car for $20,000 in 2010. The value of the car depreciates 7.2% each year. How much will the car be worth in 2016?"

This is exponential decay:
$$
A = P(1 - r)^t
$$
- $ P = 20,000 $
- $ r = 0.072 $
- $ t = 2016 - 2010 = 6 $ years

$$
A = 20,000 (1 - 0.072)^6 = 20,000 (0.928)^6
$$

Calculate:
$$
(0.928)^6 \approx 0.684
$$
$$
A \approx 20,000 \times 0.684 = 13,680
$$

✔ Answer: Approximately $13,680

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Problem 3:

"Melissa was making a bracelet, as shown in the table. The number of beads used increased each day. If the pattern continues, how many beads will she have used in total after 10 days?"

But no table is provided. However, since it mentions "as shown in the table", and it's about a pattern, likely this is geometric or arithmetic growth.

Assuming a common type of problem: suppose she uses 2 beads on Day 1, then doubles each day (geometric), or adds a constant number (arithmetic).

Since the table isn't here, let’s assume a typical geometric sequence: e.g., 2, 4, 8, 16, ... doubling each day.

Then total beads after 10 days:
$$
\text{Sum} = 2 + 4 + 8 + \ldots + 2^{10} = 2(2^{10} - 1)/(2 - 1) = 2^{11} - 2 = 2048 - 2 = 2046
$$

But without the actual table, we cannot be sure.

⚠️ Please provide the table data for an accurate answer.

However, if the problem says something like “each day she uses twice as many as the previous day”, starting from 1 bead:

Then total = $ 1 + 2 + 4 + \ldots + 2^9 = 2^{10} - 1 = 1023 $

But again — need more info.

Let me know the values in the table.

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Problem 4:

"Zac has $12,000 in a retirement fund that earns interest compounded annually at a rate of 3.8%. How much will be in his retirement fund in 15 years?"

Use compound interest:
$$
A = P(1 + r)^t
$$
- $ P = 12,000 $
- $ r = 0.038 $
- $ t = 15 $

$$
A = 12,000 (1 + 0.038)^{15} = 12,000 (1.038)^{15}
$$

Calculate:
$$
(1.038)^{15} \approx 1.807
$$
$$
A \approx 12,000 \times 1.807 = 21,684
$$

✔ Answer: Approximately $21,684

---

Problem 5:

"Lana has $9,000 in savings, earning a rate of 4.8% interest compounded continuously. How much money will be in the account in 5 years?"

Use continuous compounding:
$$
A = Pe^{rt}
$$
- $ P = 9,000 $
- $ r = 0.048 $
- $ t = 5 $
- $ e \approx 2.71828 $

$$
A = 9,000 \cdot e^{0.048 \cdot 5} = 9,000 \cdot e^{0.24}
$$

$$
e^{0.24} \approx 1.2712
$$
$$
A \approx 9,000 \times 1.2712 = 11,440.80
$$

✔ Answer: Approximately $11,440.80

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Problem 6:

"If your car loan company is making you pay interest on your loan at a rate of 4.5% per year, compounded monthly, how much will you owe after 5 years if you borrow $15,000?"

This is another compound interest problem.

$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
- $ P = 15,000 $
- $ r = 0.045 $
- $ n = 12 $
- $ t = 5 $

$$
A = 15,000 \left(1 + \frac{0.045}{12}\right)^{12 \cdot 5} = 15,000 (1 + 0.00375)^{60}
$$
$$
= 15,000 (1.00375)^{60}
$$

$$
(1.00375)^{60} \approx 1.243
$$
$$
A \approx 15,000 \times 1.243 = 18,645
$$

✔ Answer: Approximately $18,645

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $912.72 |
| 2 | $13,680 |
| 3 | Need table data (cannot solve without it) |
| 4 | $21,684 |
| 5 | $11,440.80 |
| 6 | $18,645 |

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📌 Note: For Problem 3, please provide the table showing the number of beads used each day so I can give an exact answer.

Let me know if you'd like help understanding any of the formulas or steps!