OK ILL SAY IT PEOPLE WANT MONEY TO ANSWER PROBLEMS HERE If the mean of six nu m b e rs is 48, does one
1 answer:
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Answer:
Step-by-step explanation:
Domain of f: "What values of x can we plug into this equation?" This makes sense for all real numbers so the domain is
Range of f: "What values of f(x) can we get out of the function?" From the graph we see we can get any real number greater than 0 out of the function by choosing a suitable x-value in the domain. The range is therefore .
Continuity: Since the graph is one, unbroken curve (i.e. a curve that can be drawn in one movement without taking your pen off the paper). We see that "roughly speaking" the function is continuous.
Increasing or decreasing behaviour: For all x in the domain, as x increases, f(x) decreases. This means the function exhibits decreasing behaviour.
Symmetry: It is clear to see the graph of f(x) has no symmetry.
Boundedness: Looking at the graph we see it is unbounded above as when we choose negative values, the graph of f(x) explodes upwards exponentially. Choose a value of x, plug it in, next choose (x-1), plug this in and we observe for all x in the domain.
The function is however bounded below by 0: no value of x in the domain exists which satisfies .
Extrema: As far as I can tell, there are no turning points on the curve. (Is this what you mean by extrema?)
Asymptotes: Contrary to the curve's appearance, there are no vertical asymtotes for this curve. The negative-x portion of the curve is just growing so quickly it appears to look like an asymptote. There is a value of f(x) for all x<0. There is however a horizontal asymtote: .
End behaviour: As . As
Answer:
a) starting height: 5.5 ft
b) hang time: 5.562 seconds
c) maximum height: 126.5 ft
d) time to maximum height: 2.75 seconds
Step-by-step explanation:
a) The starting height is the height at t=0.
h(0) = -16·0 +88·0 +5.5
h(0) = 5.5
The starting height is 5.5 feet.
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b) The ball is in the air between t=0 and the non-zero time when h(t) = 0. We can find the latter by solving ...
-16t^2 + 8t +5.5 = 0
t^2 -(11/2)t = 5.5/16 . . . . . subtract 5.5, then divide by -16
t^2 -(11/2)t +(11/4)^2 = (5.5/16) +(11/4)^2 . . . . complete the square
(t -11/4)^2 = 126.5/16 . . . . . . . . . . . . . . . . . . . . call this [eq1] for later use
t -11/4 = √7.90625
t = 2.75 +√7.90625 ≈ 5.562
The ball will be in the air about 5.562 seconds.
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c) If we multiply [eq1] above by -16 and add the constant on the right, we get the vertex form of the height equation:
h(t) = -16(t -11/4) +126.5
The vertex at (2.75, 126.5) tells us ...
The maximum height of the ball is 126.5 feet.
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d) That same vertex point tells us ...
The maximum height will be reached at t = 2.75 seconds.
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If you really need answers fast, a graphing calculator can give them to you in very short order (less than a minute).
