2. Is the vertex of the graph a maximum or a minimum?
maximum
minimum
neither
both
3.A ball is thrown into the air with an initial upward velocity of 60 ft/s. Its height h in feet after t seconds is given by the function h = –16t² + 60t + 6. What will the height be at t = 3 seconds?
35 feet
40 feet
42 feet
45 feet
4.Solve x² – 121 = 0
There is no solution.
–11
11
11, –11
5. Solve by factoring. m² + 8m + 7 = 0
8, 7
–7, 1
–7, –1
7, 1
6. Solve by factoring. n² + 2n – 24 = 0
–12, 2
–2, 12
–6, 4
–4, 6
7. Solve by factoring. 2p2 = 5p
(1 point)
Responses

0, 2/5
0, 5/2
0, -2/5
0, -5/2
0, 5/2
2. Is the vertex of the graph a maximum or a minimum?
maximum
minimum
neither
both
3.A ball is thrown into the air with an initial upward velocity of 60 ft/s. Its height h in feet after t seconds is given by the function h = –16t² + 60t + 6. What will the height be at t = 3 seconds?
35 feet
40 feet
42 feet
45 feet
4.Solve x² – 121 = 0
There is no solution.
–11
11
11, –11
5. Solve by factoring. m² + 8m + 7 = 0
8, 7
–7, 1
–7, –1
7, 1
6. Solve by factoring. n² + 2n – 24 = 0
–12, 2
–2, 12
–6, 4
–4, 6
2. The vertex of the graph is a minimum.
3. The height of the ball at t = 3 seconds is 42 feet.
4. The solutions to x² - 121 = 0 are x = 11 and x = -11.
5. The solutions to m² + 8m + 7 = 0 are m = -7 and m = -1.
6. The solutions to n² + 2n - 24 = 0 are n = 4 and n = -6.
ne more rectangular-shaped piece of metal siding needs to be cut to cover the exterior of a pole barn. The area of the piece is 30 ft². The length is 1 less than 3 times the width. How wide should the metal piece be? Round to the nearest hundredth of a foot.
(1 point)
Responses

3.33 ft
3.33 ft

4.3 ft
4.3 ft

1 ft
1 ft

30 ft
30 ft
To solve this problem, we need to use the given information that the area of the metal piece is 30 ft² and the length is 1 less than 3 times the width.
Let's represent the width as x:
The length would then be (3x - 1).
The area of a rectangle is given by the formula: Area = Length * Width
So we have: 30 = (3x - 1) * x
Expanding and rearranging the equation gives us: 3x^2 - x - 30 = 0
Now we can solve this quadratic equation for x:
x = [1 ± sqrt(1 + 12*30)] / 6
x = [1 ± sqrt(361)] / 6
x = [1 ± 19] / 6
The possible values for x are:
x = (1 + 19) / 6 = 20 / 6 ≈ 3.33 ft
x = (1 - 19) / 6 = -18 / 6 = -3 (Negative value is not applicable in the context of width)
Therefore, the width of the metal piece should be approximately 3.33 ft, rounded to the nearest hundredth of a foot.