2(5-6)<8f-130
(2x5)-(2x6)<8f-130
10-12<8f-130
-2<8f-130
-2+130<8f-130+130
128<8f
128/8<8f/8
16<f
f>16
Answer:
I dont know
Step-by-step explanation:
I need to know what questions you need done because all of them is a lot.
1. a
2. d
3.b
4.c
msg me if u need explanations
Answer: 40.9
Step-by-step explanation:
Given
angle given is
Using trigonometry
Using cosine formula
Given that the equation to find the height of the firework is
h(t) = at² + vt + h₀
with a = -16 ft/s² and v = 128 ft/s. In addition, since the firework starts from the ground, then the initial height, h₀, is equal to 0. Substituting these values, we have
h(t) = -16t² + 128t + 0 = -16t² + 128t
Seeing that h(t) is a quadratic function, then it forms a parabola. To find its maximum height, we can compute for the parabola's vertex.
To find the vertex's x-coordinate, we can use
t = -b/2a = (-128)/(2 · -16) = -128/-32 = 4
Since, it takes 4 seconds for the firework to reach its maximum height, then the maximum height it reaches is equal to h(4). Hence, we have
h(4) = -16(4)² + 128(4) = -16(16) + 512 = 256
Hence, the highest that the firework can reach is equal to 256 ft.
Answer: A. 256 ft