The zeros are the values of t for which f(t) = 0.
i.e. <span>-16t^2 + 96t = 0
16</span>t^2 - 96t = 16t(t - 6)
16t = 0 or t - 6 = 0
t = 0 or t = 6
Therefore, the zeros are 0, 6
The time taken for the ball to hit the ground is the value of t when f(t) = 0.
i.e. t = 6.
The question is as following:
<span>How does the graph of g(x)=1/x-5+2 compare to the graph of the parent function f(x)=1/x ?
</span>
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Solution :
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The given function ⇒

The parent function of the given function ⇒

By graphing both of the equations as shown in the attached figure.
So, we can conclude the following
<span>g(x) is shifted 5 units right and 2 units up from f(x).
</span>
So, the correct answer is option 2<span />
Step-by-step explanation:
1. C = the graphing is slowly increasing, then he walks across the top so it is flat
2. D = it started out slow and gradually got higher, which is what the story said
3. H = he had to stop walking, and there is a part of no movement in the graph H
4.
Event: Probability: A. Too much enamel 0.18 B. Too little enamel 0.24 C. Uneven application 0.33 D. No defects noted 0.47
let P(AC) = x, P(BC) = y, then P(A) + P(B) + P(C) - (x+y) = 1-0.47 = 0.53 x+y = 0.22
3. The probability of paint defects that results to <span>an improper amount of paint and uneven application? </span>
P(A U B U C) = 0.53
4. <span>the probability of a paint defect that results to</span>
<span>the proper amount of paint, but uneven application?</span>
P(C) - P(AC) - P(BC) = 0.47 - 0.22 = 0.25
A and B are disjoint so P(ABC) = 0, but you can have P(AC) and P(BC). you can't compute these separately here, but you can compute P(AC) + P(BC). By the way, P(AC) eg is just an abbreviated version of P(A∩C).
Probably $87 based on the information I was given.