Functional Relationships

lessonbunny.com

Answer Key & Explanations

Functional Relationships · Grade 8 · Worksheet 1

1. Noah is tracking the height of a bamboo plant over several days. On Day 0, the plant is 10 cm tall. From Day 0 to Day 4, the plant grows at a steady rate of 3 cm per day. From Day 4 to Day 7, the plant grows more slowly, adding only 1 cm per day. After Day 7, the plant stops growing and remains at the same height. Describe qualitatively how the height of the bamboo plant changes over time. In which intervals is the function increasing, and where is it constant? Answer: The function is increasing from Day 0 to Day 4 (steeper increase) and from Day 4 to Day 7 (less steep increase), and constant after Day 7. Solution: Identify the intervals. From Day 0 to Day 4, the height increases by 3 cm each day, so the function is increasing.
   Full step-by-step solution

   Step 1: Identify the intervals. From Day 0 to Day 4, the height increases by 3 cm each day, so the function is increasing. From Day 4 to Day 7, the height increases by 1 cm each day, so the function is still increasing but at a slower rate. From Day 7 onward, the height does not change, so the function is constant. Step 2: Describe the relationship qualitatively: The height increases over the first 7 days, with a faster increase in the first 4 days and a slower increase in the next 3 days. After Day 7, the height is constant. Answer: The function is increasing from Day 0 to Day 4 (steeper increase) and from Day 4 to Day 7 (less steep increase), and constant after Day 7.

2. Charlotte is tracking the height of a plant over 8 weeks. The height (in cm) at the end of each week is: Week 1: 4 cm, Week 2: 7 cm, Week 3: 9 cm, Week 4: 10 cm, Week 5: 10 cm, Week 6: 10 cm, Week 7: 13 cm, Week 8: 17 cm. Describe qualitatively how the height changes over time, specifying intervals where the function is increasing, decreasing, or constant. Answer: The height is increasing from Week 1 to Week 4, constant from Week 4 to Week 6, increasing from Week 6 to Week 8, and never decreasing. Solution: List the heights: Week 1: 4, Week 2: 7, Week 3: 9, Week 4: 10, Week 5: 10, Week 6: 10, Week 7: 13, Week 8: 17.
   Full step-by-step solution

   Step 1: List the heights: Week 1: 4, Week 2: 7, Week 3: 9, Week 4: 10, Week 5: 10, Week 6: 10, Week 7: 13, Week 8: 17. Step 2: Compare consecutive weeks: - Week 1 to Week 2: 4 to 7 (increase) - Week 2 to Week 3: 7 to 9 (increase) - Week 3 to Week 4: 9 to 10 (increase) - Week 4 to Week 5: 10 to 10 (constant) - Week 5 to Week 6: 10 to 10 (constant) - Week 6 to Week 7: 10 to 13 (increase) - Week 7 to Week 8: 13 to 17 (increase) Step 3: Identify intervals: - Increasing: Week 1 to Week 4 (weeks 1-4) - Constant: Week 4 to Week 6 (weeks 4-6) - Increasing: Week 6 to Week 8 (weeks 6-8) - Decreasing: None. The answer: The height is increasing from Week 1 to Week 4, constant from Week 4 to Week 6, increasing from Week 6 to Week 8, and never decreasing.

3. Liam is designing a rectangular garden with a length that is 3 feet more than twice its width. The total 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 the width of the garden be \( w \) feet. The length is 3 feet more than twice the width, so: length \( l = 2w + 3 \). Area of a rectangle = length × width.
   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 a 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** \[ 2w^2 + 3w = 65 \] Subtract 65 from both sides: \[ 2w^2 + 3w - 65 = 0 \] --- **Step 4: Solve the quadratic equation** We can use factoring: We need two numbers whose product is \( 2 \times (-65) = -130 \) and whose sum is \( 3 \). The numbers are 13 and -10 (because \( 13 \times -10 = -130 \) and \( 13 + (-10) = 3 \)). Rewrite \( 3w \) as \( 13w - 10w \): \[ 2w^2 + 13w - 10w - 65 = 0 \] Group terms: \[ (2w^2 + 13w) + (-10w - 65) = 0 \] Factor each group: \[ w(2w + 13) - 5(2w + 13) = 0 \] Factor out \( (2w + 13) \): \[ (2w + 13)(w - 5) = 0 \] --- **Step 5: Find possible values of w** Set each factor to zero: \[ 2w + 13 = 0 \quad \Rightarrow \quad w = -6.5 \quad \text{(not valid, width can't be negative)} \] \[ w - 5 = 0 \quad \Rightarrow \quad w = 5 \] So width \( w = 5 \) feet. --- **Step 6: Find length** \[ l = 2w + 3 = 2(5) + 3 = 10 + 3 = 13 \] So length \( l = 13 \) feet. --- **Step 7: Check** Area = \( 13 \times 5 = 65 \) square feet. Length \( 13 \) is indeed 3 more than twice the width \( 5 \) (since \( 2 \times 5 + 3 = 13 \)). --- **Final answer:** width = 5 feet, length = 13 feet

4. (2.5 × 10³) × (4.0 × 10⁻²) ÷ (5.0 × 10¹) = ? Answer: 2 Solution: First multiply the coefficients: 2.5 × 4.0 = 10.0 Add the exponents for multiplication: 3 + (-2) = 1 So we have 10.0 × 10¹ ÷ (5.0 × 10¹) Now divide the coefficients: 10.0 ÷ 5.0 = 2.0 Subtract the exponents: 1 - 1 = 0 2.0 × 10⁰ = 2.0 × 1 = 2 The answer is 2.
   Full step-by-step solution

   Step 1: First multiply the coefficients: 2.5 × 4.0 = 10.0 Step 2: Add the exponents for multiplication: 3 + (-2) = 1 Step 3: So we have 10.0 × 10¹ ÷ (5.0 × 10¹) Step 4: Now divide the coefficients: 10.0 ÷ 5.0 = 2.0 Step 5: Subtract the exponents: 1 - 1 = 0 Step 6: 2.0 × 10⁰ = 2.0 × 1 = 2 The answer is 2.

5. (2.4 × 10³) × (1.5 × 10⁻²) ÷ (3 × 10²) = ? Answer: 0.12 Solution: Multiply the first two terms: (2.4 × 1.5) × 10^(3 + (-2)) = 3.6 × 10^1 Now divide by the third term: (3.6 ÷ 3) × 10^(1 - 2) = 1.2 × 10^(-1) Convert to standard form: 1.2 × 10^(-1) = 0.12 The answer is 0.12.
   Full step-by-step solution

   Step 1: Multiply the first two terms: (2.4 × 1.5) × 10^(3 + (-2)) = 3.6 × 10^1 Step 2: Now divide by the third term: (3.6 ÷ 3) × 10^(1 - 2) = 1.2 × 10^(-1) Step 3: Convert to standard form: 1.2 × 10^(-1) = 0.12 The answer is 0.12.

6. Sophia is tracking the temperature of a liquid as it cools. The temperature in degrees Celsius over time in minutes is given by the function T(t) = 36 - 6t, where t is the time in minutes. Describe how the temperature changes over time. Is the function increasing, decreasing, or constant? Answer: decreasing Solution: The function is T(t) = 36 - 6t. Step 2: As t increases, the term -6t becomes more negative, so T(t) gets smaller. Step 3: For example, at t = 0, T = 36; at t = 1, T = 30; at t = 2, T = 24.
   Full step-by-step solution

   Step 1: The function is T(t) = 36 - 6t. Step 2: As t increases, the term -6t becomes more negative, so T(t) gets smaller. Step 3: For example, at t = 0, T = 36; at t = 1, T = 30; at t = 2, T = 24. Step 4: The temperature drops by 6 degrees each minute. Step 5: Therefore, the function is decreasing over time. The answer is decreasing.

7. Mere is tracking the water level in a tank. The tank starts at 24 cm. Water is added at a rate of 3 cm per minute for the first 8 minutes. Then, water is drained at a rate of 2 cm per minute for the next 6 minutes. Describe qualitatively how the water level changes over time (increasing, decreasing, or constant) during each interval. Answer: Increasing from 0 to 8 minutes, then decreasing from 8 to 14 minutes. Solution: From 0 to 8 minutes, water is added at 3 cm per minute. Since water is being added, the water level is increasing. The rate is positive, so the function is increasing during this interval.
   Full step-by-step solution

   Step 1: From 0 to 8 minutes, water is added at 3 cm per minute. Since water is being added, the water level is increasing. The rate is positive, so the function is increasing during this interval. Step 2: From 8 to 14 minutes, water is drained at 2 cm per minute. Since water is being removed, the water level is decreasing. The rate is negative, so the function is decreasing during this interval. Step 3: There is no interval where the water level is constant because the rate is never zero (either adding or draining). The answer is: Increasing from 0 to 8 minutes, then decreasing from 8 to 14 minutes.