Question
U3L12: SAMPLE WORK(1pt for correct answer and 1 pt for showing work = 10pts total) MUST SHOW HANDWRITTEN WORK FOR CREDIT...not just handwritten answer.
(-6.8) x (3.4) = ?
Multiply -2 ⅕ by -8. Write as simplifed mixed number
Lake Acid has a depth of 120ft below the surfacel. Lake Placid is ⅓ as deep. How deep is Lake Placid? Use appropriate integer to account for elevation.
Find the quotient of 8 ¼ and 2 ½. Simplified fraction
A tank holds 500 gallons of water. It drains 24 gallons of water in 6 minutes. What is the change in water in 1 minute? (remember integer representation is necessary)
(-6.8) x (3.4) = ?
Multiply -2 ⅕ by -8. Write as simplifed mixed number
Lake Acid has a depth of 120ft below the surfacel. Lake Placid is ⅓ as deep. How deep is Lake Placid? Use appropriate integer to account for elevation.
Find the quotient of 8 ¼ and 2 ½. Simplified fraction
A tank holds 500 gallons of water. It drains 24 gallons of water in 6 minutes. What is the change in water in 1 minute? (remember integer representation is necessary)
Answers
GPT-4o mini
Sure! Here are the solutions to each problem, including the handwritten work along with details for credit.
### 1. Multiply (-6.8) x (3.4)
**Solution:**
\[
-6.8 \times 3.4
\]
**Step 1: Multiply the absolute values.**
\[
6.8 \times 3.4 = 23.12
\]
**Step 2: Add the negative sign.**
\[
-6.8 \times 3.4 = -23.12
\]
**Answer:**
\[
(-6.8) \times (3.4) = -23.12
\]
---
### 2. Multiply -2 ⅕ by -8
**Solution:**
\[
-2 \frac{1}{5} \times -8
\]
**Step 1: Convert mixed number to improper fraction.**
\[
-2 \frac{1}{5} = -\frac{11}{5}
\]
**Step 2: Multiply -\(\frac{11}{5}\) by -8.**
\[
-\frac{11}{5} \times -8 = \frac{11 \times 8}{5} = \frac{88}{5}
\]
**Step 3: Convert to a mixed number.**
\[
\frac{88}{5} = 17 \frac{3}{5}
\]
**Answer:**
\[
-2 \frac{1}{5} \times -8 = 17 \frac{3}{5}
\]
---
### 3. Depth of Lake Placid
**Solution:**
\[
\text{Depth of Lake Acid} = -120 \text{ ft}
\]
\[
\text{Depth of Lake Placid} = \frac{1}{3} \text{ of Lake Acid}
\]
**Step 1: Calculate the depth.**
\[
\text{Depth of Lake Placid} = \frac{1}{3} \times (-120) = -40 \text{ ft}
\]
**Answer:**
\[
\text{Depth of Lake Placid} = -40 \text{ ft}
\]
---
### 4. Find the quotient of \(8 \frac{1}{4}\) and \(2 \frac{1}{2}\)
**Solution:**
\[
\frac{8 \frac{1}{4}}{2 \frac{1}{2}}
\]
**Step 1: Convert mixed numbers to improper fractions.**
\[
8 \frac{1}{4} = \frac{33}{4}, \quad 2 \frac{1}{2} = \frac{5}{2}
\]
**Step 2: Divide the fractions.**
\[
\frac{33}{4} \div \frac{5}{2} = \frac{33}{4} \times \frac{2}{5} = \frac{66}{20}
\]
**Step 3: Simplify the fraction.**
\[
\frac{66}{20} = \frac{33}{10} = 3 \frac{3}{10}
\]
**Answer:**
\[
\frac{8 \frac{1}{4}}{2 \frac{1}{2}} = 3 \frac{3}{10}
\]
---
### 5. Change in water in 1 minute
**Solution:**
\[
\text{Total water} = 500 \text{ gallons}
\]
\[
\text{Water drained in 6 minutes} = 24 \text{ gallons}
\]
**Step 1: Calculate the rate of drainage per minute.**
\[
\text{Rate per minute} = \frac{24}{6} = 4 \text{ gallons/minute}
\]
**Step 2: Determine change in water level.**
\[
\text{Change in water level} = -4 \text{ gallons (since it's draining)}
\]
**Answer:**
\[
\text{Change in water in 1 minute} = -4 \text{ gallons}
\]
---
### Final Notes
Each step of work is crucial for showing the full understanding of concepts, and it's important to neatly write out each calculation for proper credit.
### 1. Multiply (-6.8) x (3.4)
**Solution:**
\[
-6.8 \times 3.4
\]
**Step 1: Multiply the absolute values.**
\[
6.8 \times 3.4 = 23.12
\]
**Step 2: Add the negative sign.**
\[
-6.8 \times 3.4 = -23.12
\]
**Answer:**
\[
(-6.8) \times (3.4) = -23.12
\]
---
### 2. Multiply -2 ⅕ by -8
**Solution:**
\[
-2 \frac{1}{5} \times -8
\]
**Step 1: Convert mixed number to improper fraction.**
\[
-2 \frac{1}{5} = -\frac{11}{5}
\]
**Step 2: Multiply -\(\frac{11}{5}\) by -8.**
\[
-\frac{11}{5} \times -8 = \frac{11 \times 8}{5} = \frac{88}{5}
\]
**Step 3: Convert to a mixed number.**
\[
\frac{88}{5} = 17 \frac{3}{5}
\]
**Answer:**
\[
-2 \frac{1}{5} \times -8 = 17 \frac{3}{5}
\]
---
### 3. Depth of Lake Placid
**Solution:**
\[
\text{Depth of Lake Acid} = -120 \text{ ft}
\]
\[
\text{Depth of Lake Placid} = \frac{1}{3} \text{ of Lake Acid}
\]
**Step 1: Calculate the depth.**
\[
\text{Depth of Lake Placid} = \frac{1}{3} \times (-120) = -40 \text{ ft}
\]
**Answer:**
\[
\text{Depth of Lake Placid} = -40 \text{ ft}
\]
---
### 4. Find the quotient of \(8 \frac{1}{4}\) and \(2 \frac{1}{2}\)
**Solution:**
\[
\frac{8 \frac{1}{4}}{2 \frac{1}{2}}
\]
**Step 1: Convert mixed numbers to improper fractions.**
\[
8 \frac{1}{4} = \frac{33}{4}, \quad 2 \frac{1}{2} = \frac{5}{2}
\]
**Step 2: Divide the fractions.**
\[
\frac{33}{4} \div \frac{5}{2} = \frac{33}{4} \times \frac{2}{5} = \frac{66}{20}
\]
**Step 3: Simplify the fraction.**
\[
\frac{66}{20} = \frac{33}{10} = 3 \frac{3}{10}
\]
**Answer:**
\[
\frac{8 \frac{1}{4}}{2 \frac{1}{2}} = 3 \frac{3}{10}
\]
---
### 5. Change in water in 1 minute
**Solution:**
\[
\text{Total water} = 500 \text{ gallons}
\]
\[
\text{Water drained in 6 minutes} = 24 \text{ gallons}
\]
**Step 1: Calculate the rate of drainage per minute.**
\[
\text{Rate per minute} = \frac{24}{6} = 4 \text{ gallons/minute}
\]
**Step 2: Determine change in water level.**
\[
\text{Change in water level} = -4 \text{ gallons (since it's draining)}
\]
**Answer:**
\[
\text{Change in water in 1 minute} = -4 \text{ gallons}
\]
---
### Final Notes
Each step of work is crucial for showing the full understanding of concepts, and it's important to neatly write out each calculation for proper credit.