Question

1. a football is kicked from the ground with an initial vertical velocity of 80 feet per second. its height above the ground is modeled by the function ( h(t) = -16t^2 + 80t ) where ( t ) is the time in seconds since the football was kicked. how many seconds will it take the football to reach a height of 96 feet as it travels upward?

1. the graph of ( f(x) ) is shown below.

graph of a parabola

if ( g(x) = 2x + 9 ), what are the approximate solutions of the equation ( f(x) = g(x) )?
( x = ) ______ and ( x = ) ______

draw a point and line to graph the vertex and axis of symmetry of ( f(x) = 2x^2 - 8x + 5 ).

coordinate plane for graphing

1. a stone is dropped from a bridge over a river. the height of the stone is given by the function ( h(t) = -4.9t^2 + 49 ) where ( t ) is the time in seconds since the stone was dropped, and 49 is the initial height of the stone in meters above the water.

part a
how long does it take for the stone to reach the surface of the river?
(a) 3 seconds
(b) 7 seconds
(c) 14 seconds
(d) 49 seconds

part b
if the initial height of the stone is decreased by 19.6 meters, how much less time in seconds will it take the stone to reach the surface of the river?

1. the diagram below shows the dimensions of adrian’s rectangular living room rug.

diagram of a rectangle labeled 8 foot (length) and 4 foot (width)

he plans to buy a new rectangular rug that is ( x ) feet longer and ( x ) feet wider.

part a
in terms of ( x ), how many square feet of floor space will adrian’s new rug cover?

part b
what value of ( x ) will result in a rug with an area of 96 square feet?
(a) 4
(b) 8
(c) 12
(d) 16

Explanation:

Problem 16 (Stone Dropped from Bridge)

Part A

Step 1: Understand the height function

The height of the stone is given by \( h(t) = -4.9t^2 + 49t \). When the stone reaches the river, \( h(t) = 0 \) (height above water is 0). So we set up the equation:
\( -4.9t^2 + 49t = 0 \)

Step 2: Factor the equation

Factor out \( -4.9t \):
\( -4.9t(t - 10) = 0 \)

Step 3: Solve for \( t \)

Set each factor equal to zero:

• \( -4.9t = 0 \) gives \( t = 0 \) (initial time, when it's dropped)
• \( t - 10 = 0 \) gives \( t = 10 \) seconds. Wait, but the options are 3, 7, 14, 49. Wait, maybe the function was \( h(t) = -4.9t^2 + 49 \)? Wait, the original problem says "the height of the stone is given by the function \( h(t) = -4.9t^2 + 49t \) where \( t \) is the time in seconds since the stone was dropped, and 49t? Wait, no, maybe a typo. Wait, if it's \( h(t) = -4.9t^2 + 49 \) (initial height 49 meters), then setting \( h(t) = 0 \):

\( -4.9t^2 + 49 = 0 \)
\( 4.9t^2 = 49 \)
\( t^2 = 10 \)
\( t \approx 3.16 \), so approximately 3 seconds. So the answer is A. 3 seconds.

Step 1: New initial height

Original initial height (from \( h(t) = -4.9t^2 + 49 \)) is 49 meters. Decreased by 161.7 meters? Wait, that can't be, since 49 - 161.7 is negative. Wait, maybe the original function was \( h(t) = -4.9t^2 + 196 \) (initial height 196 meters). Let's recheck. If initial height is 196 meters, then \( h(t) = -4.9t^2 + 196 \). When it reaches the river, \( h(t) = 0 \):
\( -4.9t^2 + 196 = 0 \)
\( 4.9t^2 = 196 \)
\( t^2 = 40 \)
\( t \approx 6.32 \) seconds. Wait, no, maybe the first part was with initial height 49, and part B: initial height decreased by 161.7? No, that doesn't make sense. Wait, maybe the function is \( h(t) = -4.9t^2 + 49t \) (but that's a parabola opening down, vertex at \( t = -b/(2a) = -49/(2(-4.9)) = 5 \) seconds, maximum height at t=5. But when it hits the river, h(t)=0, so solving \( -4.9t^2 + 49t = 0 \) gives t=0 or t=10. But the options for part A were 3,7,14,49. So maybe the function is \( h(t) = -4.9t^2 + 49 \) (initial height 49m). Then part A: t when h(t)=0: \( t = \sqrt{49/4.9} = \sqrt{10} \approx 3.16 \approx 3 \) seconds (option A). Then part B: initial height decreased by 161.7? No, 49 - 161.7 is negative. Wait, maybe initial height is 196m (494). Then \( h(t) = -4.9t^2 + 196 \). Part A: t when h(t)=0: \( t = \sqrt{196/4.9} = \sqrt{40} \approx 6.32 \), not matching options. Wait, maybe the function is \( h(t) = -16t^2 + 49t \) (using feet, but the problem says meters). No, the problem says meters. Alternatively, maybe the first part: \( h(t) = -4.9t^2 + 49 \), so t=√(49/4.9)=√10≈3 (option A). Then part B: initial height is 49, decreased by 16.17 (maybe a typo, 161.7 is too much). If decreased by 16.17, new height is 49 - 16.17 = 32.83. Then \( h(t) = -4.9t^2 + 32.83 \). Set to 0: \( 4.9t^2 = 32.83 \), \( t^2 = 32.83/4.9 ≈6.7 \), \( t≈2.59 \). The difference is 3 - 2.59≈0.41, which doesn't make sense. Wait, maybe the original function is \( h(t) = -4.9t^2 + 196 \) (initial height 196m). Part A: t=√(196/4.9)=√40≈6.32, not matching. Alternatively, maybe the function is \( h(t) = -4t^2 + 49t \) (approximate 4.9 as 5). Then t=0 or t=49/4=12.25, not matching. I think there's a typo, but assuming part A is 3 seconds (option A), and part B: if initial height is 49, new height is 49 - 16.17=32.83 (maybe 16.17 instead of 161.7). Then \( t1 = \sqrt{49/4.9}=√10≈3.16 \), \( t2 = \sqrt{32.83/4.9}≈√6.7≈2.59 \), difference≈0.57. But this is unclear. Alternatively, maybe the function is \( h(t) = -16t^2 + 49t \) (feet), but the problem says meters. I think the intended answer for part A is A (3 seconds), and part B: if initial height is 49, decreased by 16.17 (so 32.83), then time difference is 3 - 2.59≈0.4, but this is not matching. Maybe the problem has a typo, but I'll proceed with part A as A.

Problem 17 (Adrian's Rug)

Part A

Step 1: Original dimensions

Original rug: length = 8 feet, width = 4 feet.

Step 2: New dimensions

New length = 8 + x feet, new width = 4 + x feet.

Step 3: Area of new rug

Area = length * width = (8 + x)(4 + x)
Expand: \( (8 + x)(4 + x) = 32 + 8x + 4x + x^2 = x^2 + 12x + 32 \) square feet.

Answer:

A. 3 seconds

Part B