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Algebra 1

8.1.5 Practice

Algebra 18.1.5 Practice
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Complete the following questions to practice the skills you have learned in this lesson.

Use the following scenario to answer questions 1 – 3.

A water balloon is launched by a catapult into the air. The water balloon is launched from 4 feet above the ground, with an initial vertical velocity of 40 feet per second.

The function h ( t ) = − 16 t 2 + 40 t + 4 models the height of the water balloon over the ground, in feet, t seconds after launch.

A DOWNWARD PARABOLA. X-AXIS LABEL IS TIME IN SECONDS AND Y-AXIS LABEL IS HEIGHT IN FEET. THE GRAPH PASSES THROUGH THE POINTS (0, 4), (0.5, 20), (1, 28), (1.5, 28), (2, 20), AND (2.5, 4).

  1. What is the height of the water balloon 1 second after launch?
  1. 34 feet
  2. 28 feet
  3. 22 feet
  4. 18 feet
  1. Will the water balloon still be in the air 2 seconds after the launch?
  1. No
  2. Yes
  1. Using the graph, estimate when the water balloon hit the ground.
  1. 2.6 seconds
  2. 2.1 seconds
  3. 1.8 seconds
  4. 1.2 seconds

Use the following scenario to answer questions 4 – 5.

Suppose you want to frame a different picture that measures 12 inches by 9 inches. The frame is created from 46 square inches of framing material. The frame must be uniform in width, and all of the framing material must be used.
  1. Which of the following is true about the situation described?
  1. The perimeter of the framed picture is greater than 42 inches.
  2. The area of the framed picture is 148 square inches.
  3. The perimeter of the framed picture is 42 inches.
  4. The perimeter of the picture is 40 inches.
  1. What is the total area of the framed picture?
  1. 176 square inches
  2. 154 square inches
  3. 108 square inches
  4. 46 square inches

Use the following scenario to answer questions 6 – 8.

Suppose you want to frame a different picture that measures 9 inches by 8 inches. The frame is created from 38 square inches of framing material. The frame must be uniform in width, and all of the framing material must be used. Let x represent the thickness of the frame.
  1. What is the total area of the framed picture?
  1. 134 square inches
  2. 110 square inches
  3. 72 square inches
  4. 34 square inches
  1. Write an equation to represent the relationship between the measurements of the picture and of the frame and the area of the framed picture.
  1. 4 x 2 + 34 x + 72 = 110
  2. x 2 + 34 x + 72 = 110
  3. 4 x 2 + 34 x = 72
  4. x 2 + 34 x = 38
  1. What does the solution to the equation in the previous question represent in this situation?
  1. The solution represents the total area of the framed picture.
  2. The solution represents the perimeter of the frame when all of the framing material is used.
  3. The solution represents the thickness of the frame when all of the framing material is used.
  4. The solution represents the area of the framing material that should be used.

Use the following scenario to answer questions 9 – 10.

A framed picture has a total area y , in square inches. The thickness of the frame is represented by x , in inches. The equation y = ( 2 x + 5 ) ( 2 x + 7 ) relates these two variables.
  1. What are the width and length of the picture without the frame?
  1. 5 inches and 7 inches
  2. 2 inches and 7 inches
  3. 2 inches and 5 inches
  4. 2 inches and 2 inches
  1. Which of the following represents the area of the frame?
  1. 35
  2. x + 35
  3. x + y
  4. y − 35
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