Answer:
Step-by-step explanation:
To solve this problem you need the function
h(t) = -16t2 + v0t + h0
where t = time
v0 is the initial velocity, which in our case is 0
h0 = initial height, which in our case is 256
h(t) = 0 since we want to know when the ball will hit the ground.
0 = -16 t2 + 256
And we can solve for t
If we rearrange the terms we see that this is a difference of 2 squares
0 = 256 - 16t2
0 = (16-4t)(16+4t)
Setting each factor = 0
16-4t=0 16+4t=4
t = 4 t = -4
The second solution is discarded as time cannot be negative.
So the ball will hit the ground in 4 seconds.
Simplifying h(x) gives
h(x) = (x² - 3x - 4) / (x + 2)
h(x) = ((x² + 4x + 4) - 4x - 4 - 3x - 4) / (x + 2)
h(x) = ((x + 2)² - 7x - 8) / (x + 2)
h(x) = ((x + 2)² - 7 (x + 2) - 14 - 8) / (x + 2)
h(x) = ((x + 2)² - 7 (x + 2) - 22) / (x + 2)
h(x) = (x + 2) - 7 - 22/(x + 2)
h(x) = x - 5 - 22/(x + 2)
An oblique asymptote of h(x) is a linear function p(x) = ax + b such that

In the simplified form of h(x), taking the limit as x gets arbitrarily large, we obviously have -22/(x + 2) converging to 0, while x - 5 approaches either +∞ or -∞. If we let p(x) = x - 5, however, we do have h(x) - p(x) approaching 0. So the oblique asymptote is the line y = x - 5.
Answer:
8
Step-by-step explanation:
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Answer:
if there is equivalent ratio cookies should be 48