Functional Relationships

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Functional Relationships · Grade 8 · Worksheet 3

1. Mason is observing a hot air balloon ascending and descending over time. He records its height above the ground every 10 minutes. From 0 to 20 minutes, the balloon rises steadily from 0 meters to 40 meters. Then, from 20 to 30 minutes, it stays at a constant height of 40 meters. From 30 to 50 minutes, it descends steadily back to the ground (0 meters). Describe qualitatively how the height of the balloon changes over time in each of these three time intervals: 0–20 minutes, 20–30 minutes, and 30–50 minutes. Answer: From 0 to 20 minutes, the height is increasing. From 20 to 30 minutes, the height is constant. From 30 to 50 minutes, the height is decreasing. Solution: Look at the time interval from 0 to 20 minutes. The problem says the balloon 'rises steadily' from 0 to 40 meters. This means the height is going up over time, so the relationship is increasing.
   Full step-by-step solution

   Step 1: Look at the time interval from 0 to 20 minutes. The problem says the balloon 'rises steadily' from 0 to 40 meters. This means the height is going up over time, so the relationship is increasing. Step 2: Look at the time interval from 20 to 30 minutes. The problem says the balloon 'stays at a constant height' of 40 meters. This means the height is not changing, so the relationship is constant. Step 3: Look at the time interval from 30 to 50 minutes. The problem says the balloon 'descends steadily' back to the ground. This means the height is going down over time, so the relationship is decreasing. Final answer: From 0 to 20 minutes, the height is increasing. From 20 to 30 minutes, the height is constant. From 30 to 50 minutes, the height is decreasing.

2. Describe how the function y = 3x² - 5 changes as x increases from -3 to 0, and then from 0 to 3. Is the function increasing, decreasing, or constant in each interval? Answer: From x = -3 to 0, the function is decreasing. From x = 0 to 3, the function is increasing. Solution: Recognize that y = 3x² - 5 is a quadratic function with a positive coefficient (3) for x². This means its graph is a parabola opening upward, with a minimum point at the vertex.
   Full step-by-step solution

   Step 1: Recognize that y = 3x² - 5 is a quadratic function with a positive coefficient (3) for x². This means its graph is a parabola opening upward, with a minimum point at the vertex. Step 2: The vertex of y = 3x² - 5 occurs at x = 0 (since there is no x term). At x = 0, y = 3(0)² - 5 = -5. Step 3: Consider the interval from x = -3 to 0. As x increases from -3 to 0, the value of x² decreases (since (-3)² = 9, (-2)² = 4, (-1)² = 1, 0² = 0). So 3x² decreases, and therefore y = 3x² - 5 also decreases. The function is decreasing on this interval. Step 4: Consider the interval from x = 0 to 3. As x increases from 0 to 3, the value of x² increases (0² = 0, 1² = 1, 2² = 4, 3² = 9). So 3x² increases, and therefore y = 3x² - 5 also increases. The function is increasing on this interval. Step 5: Conclusion: From x = -3 to 0, the function is decreasing. From x = 0 to 3, the function is increasing.

3. Noah is tracking the water level in a large tank as it is being drained for cleaning. The tank starts with 1800 gallons of water. He opens the drain valve, and water flows out at a constant rate of 12 gallons per minute. Describe qualitatively how the amount of water in the tank changes over time from the moment the valve is opened until the tank is empty. Be sure to state whether the relationship is increasing, decreasing, or constant, and explain why. Answer: The amount of water in the tank is decreasing at a constant rate over time until it reaches zero. Solution: At the start, the tank has 1800 gallons of water. As time passes, water flows out at 12 gallons per minute, so the total amount of water in the tank decreases.
   Full step-by-step solution

   Step 1: At the start, the tank has 1800 gallons of water. As time passes, water flows out at 12 gallons per minute, so the total amount of water in the tank decreases. Step 2: The rate of decrease is constant (12 gallons every minute), so the relationship is decreasing at a constant rate. Step 3: This continues until the tank is empty (0 gallons). Therefore, the relationship is decreasing and constant (linear decrease).

4. (2.4 × 10³) × (3.0 × 10⁻²) ÷ (4.0 × 10¹) = ? Answer: 1.8 Solution: Multiply the first two terms: (2.4 × 3.0) × 10^(3 + (-2)) = 7.2 × 10¹ Now divide by the third term: (7.2 ÷ 4.0) × 10^(1 - 1) = 1.8 × 10⁰ Since 10⁰ = 1, the final answer is 1.8 The answer is 1.8.
   Full step-by-step solution

   Step 1: Multiply the first two terms: (2.4 × 3.0) × 10^(3 + (-2)) = 7.2 × 10¹ Step 2: Now divide by the third term: (7.2 ÷ 4.0) × 10^(1 - 1) = 1.8 × 10⁰ Step 3: Since 10⁰ = 1, the final answer is 1.8 The answer is 1.8.

5. Emma is training for a cycling race. She records her distance traveled over time during a 90-minute training ride. During the first 30 minutes, she cycles at a steady pace and covers 12 miles. For the next 15 minutes, she stops to fix a flat tire and does not move. Then, for the next 45 minutes, she cycles at a faster pace and covers 27 miles. Describe qualitatively how Emma's distance from her starting point changes over the entire 90-minute ride. Answer: Increasing, then constant, then increasing. Solution: For the first 30 minutes, Emma cycles at a steady pace and covers 12 miles. Since she is moving away from the start, the distance from the starting point is increasing.
   Full step-by-step solution

   Step 1: For the first 30 minutes, Emma cycles at a steady pace and covers 12 miles. Since she is moving away from the start, the distance from the starting point is increasing. Step 2: For the next 15 minutes, she stops to fix a flat tire and does not move. The distance from the starting point stays the same (constant) because she is not traveling. Step 3: For the final 45 minutes, she cycles at a faster pace and covers 27 miles. Since she is moving away from the start again, the distance from the starting point is increasing. Overall, the distance increases, then stays constant, then increases again.

6. A company is testing a new solar panel design. The power output P (in watts) of the panel varies with the temperature T (in degrees Celsius) according to the linear equation P = -0.5T + 120. At what temperature will the solar panel produce exactly 85 watts of power? Answer: 70 Solution: Start with the equation P = -0.5T + 120 Substitute P = 85 into the equation: 85 = -0.5T + 120 Subtract 120 from both sides: 85 - 120 = -0.5T Simplify: -35 = -0.5T Divide both sides by -0.5: T = (-35)/(-0.5) Calculate: T = 70 The answer is 70 degrees Celsius.
   Full step-by-step solution

   Step 1: Start with the equation P = -0.5T + 120 Step 2: Substitute P = 85 into the equation: 85 = -0.5T + 120 Step 3: Subtract 120 from both sides: 85 - 120 = -0.5T Step 4: Simplify: -35 = -0.5T Step 5: Divide both sides by -0.5: T = (-35)/(-0.5) Step 6: Calculate: T = 70 The answer is 70 degrees Celsius.

7. Liam is designing a rectangular garden with a length that is 3 feet more than twice its width. The area of the garden is 65 square feet. What are the dimensions of Liam's garden? Answer: width = 5 feet, length = 13 feet Solution: Let’s go step-by-step. Let the width of the garden be \( w \) feet. The length is 3 feet more than twice the width, so: length \( l = 2w + 3 \).
   Full step-by-step solution

   Let’s go step-by-step. --- **Step 1: Define variables** Let the width of the garden be \( w \) feet. The length is 3 feet more than twice the width, so: length \( l = 2w + 3 \). --- **Step 2: Write the area equation** Area of rectangle = length × width Given area = 65 square feet: \[ l \times w = 65 \] Substitute \( l = 2w + 3 \): \[ (2w + 3) \times w = 65 \] --- **Step 3: Expand and rearrange** \[ 2w^2 + 3w = 65 \] \[ 2w^2 + 3w - 65 = 0 \] --- **Step 4: Solve the quadratic equation** Use the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here \( a = 2 \), \( b = 3 \), \( c = -65 \). First, compute the discriminant: \[ b^2 - 4ac = 3^2 - 4(2)(-65) = 9 + 520 = 529 \] \[ \sqrt{529} = 23 \] So: \[ w = \frac{-3 \pm 23}{2 \times 2} = \frac{-3 \pm 23}{4} \] --- **Step 5: Two possible solutions** Case 1: \( w = \frac{-3 + 23}{4} = \frac{20}{4} = 5 \) Case 2: \( w = \frac{-3 - 23}{4} = \frac{-26}{4} = -6.5 \) Width cannot be negative, so \( w = 5 \). --- **Step 6: Find length** \[ l = 2w + 3 = 2(5) + 3 = 10 + 3 = 13 \] --- **Step 7: Verify** Area = \( 13 \times 5 = 65 \) square feet. Length is 3 more than twice the width: \( 2 \times 5 + 3 = 13 \). Correct. --- **Final answer:** width = 5 feet, length = 13 feet