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)

1 answer

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.