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bulgar [2K]
3 years ago
6

A bird drops a nut from a height of 441 feet. Its height, h, in feet is given by the equation h=-16t^2=441 where t is the time i

n seconds the nut has fallen. How many seconds will it take for the nut to hit the ground?
Mathematics
1 answer:
Ahat [919]3 years ago
8 0
<span> A bird drops a nut from a height of 441 feet.
Its height, h, in feet is given by the equation h=-16t^2=441
,where t is the time in seconds the nut has fallen.

The time it </span>will it take for the nut to hit the ground is given as:

t^{2} = \frac{441}{16} \\ \\ t = \frac{21}{4} \\ \\ t = 5.25 seconds 
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Basic functions sample work! Help please
frosja888 [35]

Answer:

  See the attachment.

Step-by-step explanation:

Function notation: This means the description of the curve is written in the form ...

  f(x) = <an expression that describes the curve in terms of independent variable x>

The name of the function does not have to be "f", though f, g, h are traditional for generic functions. In specific cases, the function name might be related to the value the function delivers: <em>cost(n)</em> might be the function that tells you the cost of producing n items, for example.

The functions shown (square root, reciprocal) should be shapes you have memorized. You can check the scale factor at specific values you know, for example, √4 = 2, or 1/1 = 1. Hence both functions have a scale factor of 1 (no scaling).

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The domain is the horizontal region where the function is defined. Square root is not defined for negative numbers; reciprocal is not defined for 0. Hence the descriptions of the respective domains must exclude these values.

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The range is the vertical extent of the values each function produces. The domain and range are the same for both functions shown here.

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Asymptotes are lines the function approaches but does not reach. The square root function has no such lines. (It reaches vertical at x=0.)

The reciprocal function has a vertical asymptote as x=0 (the value of x that makes the function denominator zero). And, it has a horizontal asymptote at y=0, a value that can be approached, but not reached, as x gets large in either the positive or negative directions.

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End behavior: The square root function heads off toward (∞, ∞). The reciprocal function heads toward the value defined by the horizontal asymptote: y=0 as x gets large in magnitude.

3 0
2 years ago
The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

  • A(0) = 10.
  • A(6) = 28.

Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

\displaystyle A^{\prime\prime}(t) = 6\, t - 15.

Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

  • A(1) = 15.5.

Compare these three possible maximum values of A(t) over the interval 0 \le t \le 6. Apparently, t = 6 would maximize the value of A(t)\!. That is: A(6) = 28 gives the maximum value of \! A(t) over the interval 0 \le t \le 6\!.

However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

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3 years ago
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yanalaym [24]

Answer:

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6 0
2 years ago
To divide 1.89 ÷ 0.9, we need to change it. Choose the correct problem below to show what the “new” look would be.
Arlecino [84]

Answer:

D. 18.9 ÷ 9

Step-by-step explanation:

we need to divide

1.89 ÷ 0.9 but write it in different form

so ,we need to eliminate decimal from 0.9

as 0.9*10 = 9

thus,we multiply both  1.89 and  0.9 with 10, then we will have

(1.89*10) ÷ (0.9*10)

=> 18.9 ÷ 9

Thus, based on above calculation new look would be D. 18.9 ÷ 9

8 0
3 years ago
What value of c is the rational expression below equal to zero? x-4/x-6
RSB [31]

Answer: x=4

Steps: (I am assuming your question has a typo in it and by "c" you meant "x")

The rational expression equals zero

\frac{x-4}{x-6}=0

only when the numerator equals 0 (the denominator cannot ever be zero):

x-4 = 0 \\x = 4

and that happens only for x=4


3 0
3 years ago
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