What Did The Pelican Say When It Finished Shopping

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nezar?!?! 1m ago Name 3.5 Puzzle Time What Did The Pelican Say When It Finished Shopping? Write the letter of each answer in the box containing the exercise number. Find the slope of the line passing through the given points. 1. (-10, -12), (-8, -8), (-6,-4), (-4, 0) 2. (-4, 2), (0, 1), (4, 0), (8, -1) 3. (-7,-7), (0, -8), (7,-9). (14, -10) 4. (-2, 2), (0, 3), (2, 4), (4,.5) 5. (2, -11), (4, -25), (6,-39), (8, -53) 6. (-11, -38), (-5, -14), (1, 10), (7, 34) Find the slope and the y-intercept of the graph of the linear equation. 7. y = -4x+6 8. y = 4 9. 4x + y = -1 10. y = 6x-4 11. -x – 4y + 8 = 0 12. 2x12y + 10 = 0 13. The local service center advertises that it charges a flat fee of $50 plus $8 per mile to tow a vehicle. The function C(x) = 8x + 50 represents the cost C (in dollars) of towing a vehicle, where x is the number of miles the vehicle is towed. Identify the slope and y-intercept. 7 2 12 mpyright © Big Ideas Leaming, LLC rights reserved. O 5 8 1 9 Send a chat Answers 1 2 M. m 6, b = -4 1. m- 10 3 U. m = N. m-4, b = -1 P. m-4, b = 6 1. m = -7 Ü� Date L. m = b = 2 0. m = 2 L. m = 4 T. m. 0, b = – = Y. m. B. m = 8, b = 50 1 7 4 5 T. m = 1, b = 2 6 13 4 11 6 : Algebra 1 Resources by Chapter

The correct answer and explanation is:

Let’s solve the given problems step by step:

1. Find the slope of the line passing through the points.

To calculate the slope mm, use the slope formula: m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}

Problem 1: Points (-10, -12), (-8, -8), (-6, -4), (-4, 0)

  • Calculate between each pair of points: m=−8−(−12)−8−(−10)=42=2m = \frac{-8 – (-12)}{-8 – (-10)} = \frac{4}{2} = 2 (The slope remains constant, m=2m = 2, because the points form a straight line.)

Problem 2: Points (-4, 2), (0, 1), (4, 0), (8, -1)

  • Calculate between each pair of points: m=1−20−(−4)=−14=−14m = \frac{1 – 2}{0 – (-4)} = \frac{-1}{4} = -\frac{1}{4} (The slope remains constant, m=−14m = -\frac{1}{4}.)

Problem 3: Points (-7, -7), (0, -8), (7, -9), (14, -10)

  • Calculate between each pair of points: m=−8−(−7)0−(−7)=−17m = \frac{-8 – (-7)}{0 – (-7)} = \frac{-1}{7} (The slope remains constant, m=−17m = -\frac{1}{7}.)

Problem 4: Points (-2, 2), (0, 3), (2, 4), (4, 5)

  • Calculate between each pair of points: m=3−20−(−2)=12m = \frac{3 – 2}{0 – (-2)} = \frac{1}{2} (The slope remains constant, m=12m = \frac{1}{2}.)

Problem 5: Points (2, -11), (4, -25), (6, -39), (8, -53)

  • Calculate between each pair of points: m=−25−(−11)4−2=−142=−7m = \frac{-25 – (-11)}{4 – 2} = \frac{-14}{2} = -7 (The slope remains constant, m=−7m = -7.)

Problem 6: Points (-11, -38), (-5, -14), (1, 10), (7, 34)

  • Calculate between each pair of points: m=−14−(−38)−5−(−11)=246=4m = \frac{-14 – (-38)}{-5 – (-11)} = \frac{24}{6} = 4 (The slope remains constant, m=4m = 4.)

2. Find the slope and y-intercept of the given equations.

Problem 7: y=−4x+6y = -4x + 6

  • Slope m=−4m = -4, y-intercept b=6b = 6.

Problem 8: y=4y = 4

  • Slope m=0m = 0, y-intercept b=4b = 4.

Problem 9: 4x+y=−14x + y = -1

  • Rewrite as y=−4x−1y = -4x – 1.
  • Slope m=−4m = -4, y-intercept b=−1b = -1.

Problem 10: y=6x−4y = 6x – 4

  • Slope m=6m = 6, y-intercept b=−4b = -4.

Problem 11: −x−4y+8=0-x – 4y + 8 = 0

  • Rewrite as y=−14x+2y = -\frac{1}{4}x + 2.
  • Slope m=−14m = -\frac{1}{4}, y-intercept b=2b = 2.

Problem 12: 2x−12y+10=02x – 12y + 10 = 0

  • Rewrite as y=16x−53y = \frac{1}{6}x – \frac{5}{3}.
  • Slope m=16m = \frac{1}{6}, y-intercept b=−53b = -\frac{5}{3}.

3. Towing Function C(x)=8x+50C(x) = 8x + 50

  • Slope m=8m = 8 (Cost per mile).
  • Y-intercept b=50b = 50 (Flat fee for towing).

Explanation

To solve problems involving slopes and y-intercepts, the formula m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1} helps find the slope, showing the rate of change between points. A constant slope across all point pairs confirms a straight line.

Linear equations in slope-intercept form y=mx+by = mx + b allow direct identification of mm (slope) and bb (y-intercept). For equations not in this form, algebraic manipulation isolates yy to convert them.

For example, in 4x+y=−14x + y = -1, rearranging yields y=−4x−1y = -4x – 1, identifying m=−4m = -4 and b=−1b = -1. In the context of real-world functions, such as the towing example C(x)=8x+50C(x) = 8x + 50, m=8m = 8 represents the variable cost per mile, while b=50b = 50 is the fixed initial cost.

Understanding slope as the measure of steepness or direction and y-intercept as the starting value of a function provides insight into data trends and relationships in various scenarios, including geometry and applied mathematics.

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