AnswersFL-1200330-Algebra 2 AThe Quadratic Formula

Modeling with Quadratic Equations Answers

10 verified answers1 views
9

When using a quadratic equation in the form y = ax2 + bx + c to model the height of a projectile (y) over time (x), which of the following is always represented by the constant term?

A
the initial height of the projectile
B
the initial velocity of the projectile
C
the time at which the projectile hits the ground
D
the maximum height of the projectile
11

Carmen is using the quadratic equation (x + 15)(x) = 100 where x represents the width of a picture frame. Which statement about the solutions x = 5 and x = –20 is true?

A
The solutions x = 5 and x = –20 are reasonable.
B
The solution x = 5 should be kept, but x = –20 is unreasonable.
C
The solution x = –20 should be kept, but x = 5 is unreasonable.
D
The solutions x = 5 and x = –20 are unreasonable.
12

A projectile is launched straight up from the ground with an initial velocity of 120 ft/s. If acceleration due to gravity is –16 ft/s2, after about how many seconds will the object reach a height of 200 ft?h(t) = at2 + vt + h0

A
1.0 s
B
1.7 s
C
2.5 s
D
8.0 s
13

The total area of two square windows is 1,025 in.2. Each side of the larger window is 5 in. longer than the sides of the smaller window. How long are the sides of the smaller window?A = s2

A
13.5 in.
B
20 in.
C
25 in.
D
31.6 in.
14

A baseball is thrown into the air from a height of 5 feet. The ball reaches a maximum height of 43.5 feet and spends a total of 3.2 seconds in the air. Which equation models the height of the baseball? Assume that acceleration due to gravity is –16 ft/s2.

A
h(t) = 16t2 + 49.64t + 5
B
h(t) = -16t2 + 5t + 49.64
C
h(t) = -16t2 + 49.64t + 5
D
h(t) = 16t2 + 5t + 49.64
15

A quarterback throws an incomplete pass. The height of the football at time t is modeled by the equation h(t) = –16t2 + 40t + 7. Rounded to the nearest tenth, the solutions to the equation when h(t) = 0 feet are –0.2 s and 2.7 s. Which solution can be eliminated and why?

A
The solution –0.2 s can be eliminated because time cannot be a negative value.
B
The solution –0.2 s can be eliminated because the pass was not thrown backward.
C
The solution 2.7 s can be eliminated because the pass was thrown backward.
D
The solution 2.7 s can be eliminated because a ball cannot be in the air for that long due to gravity.

Did you find these answers helpful?