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oksian1 [2.3K]
3 years ago
8

In the graph below, which descriptions relate to Figure c?

Mathematics
1 answer:
Bingel [31]3 years ago
6 0

im not sure what ur asking. but c is a line segment

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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.)

4 0
3 years ago
Owen received $100 for his birthday.he wants to spend 2/10 of his money on a video game.he wants to spend 55/100 of his money on
babymother [125]
An easy way to solve fraction problems like this is to make each fraction have the same denominator.  Since he has $100, the denominator will be 100.
2/10=?/100    10*10=100  1*10=20   20/100 of his money he wants to spend on video games
55/100 already has 100 as its denominator, so he wants to spend 55/100 on a skateboard.
3/10=?/100   10*10=100   3*10=100  30/100 is how much he wants to spend on comic books.
Add all of the fractions together to see how much he wants to spend and if he has enough.
20/100+55/100+30/100=105/100
He wants to spend $105, which is $5 more than he has.
8 0
2 years ago
Please help me as soon as possible
elixir [45]
You might want to see what other people say but I think that it is 40
5 0
3 years ago
Read 2 more answers
Help Please.<br><br> Let f(x)=186+5e−0.1x . What is f(6) ?
larisa86 [58]

The answer is 186+5e-0.6

7 0
3 years ago
A piece of string is 325 centimeters long. You need the string to be 114 meters long. How many centimeters should you cut off?
sertanlavr [38]

There are 100 centimeters in 1 meter.1 and 1/4 meters can also be written as 1.25, since 1/4 = 0.25.To find the answer, convert 1.25 meters to centimeters.
1 meter = 100 centimeters1.25 meters = x centimeters1.25 is 125% of 1. In other words, there is a 25% increase.
Now apply this increase to the centimeters side.125% = 1.251.25 • 100 = 125
So 1.25 meters (1 and 1/4 meters) equals 125 centimeters.
The string is 325 centimeters.In order for the string to be 125 centimeters, you need to cut some off.To find how much you need to cut off, subtract 125 from 325.325 - 125 = 200
Answer: In order for your string to be 1 and 1/4 meters, you must cut 200 centimeters off the original 325 centimeter long string.
5 0
2 years ago
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