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Ivanshal [37]
2 years ago
5

The sum of 4 and a number is at least 15

Mathematics
1 answer:
NeTakaya2 years ago
3 0
The answer should be 4+n (greater than or equal to symbol >_) 15
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The answer is 3 and 7/8.
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Help me with this please its hard and i dont get it
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Please help! Thank you! xx
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4 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.)

4 0
3 years ago
Each of these labelled bags contains two apples.
Dahasolnce [82]

Step-by-step explanation:

all bags are mislabeled. that means, whatever is written on either of the 3 bags, is wrong.

so, the bag labeled "green" is NOT the one with 2 green apples. the one labeled "red" is NOT the one with 2 red apples.

and - my favorite - the one labeled "red/green" is NOT the one with the mixture.

so, we pick one apple from the bag that is currently (mis-)labeled as "red/green".

as it cannot be the mixed bag, it must be one of the pure color bags.

whatever color shows up, we know this is the bag with 2 apples of that color.

then the bag that was labeled with that color must be the other pure color (if it were the mixed bag, it would mean that the third bag was labeled correctly, which is not the case). and then the third bag is the mixed one.

for example :

we get a red apple in our first pick.

so, the bag, that was originally (but wrongly) labeled as "red/green" gets the label "red".

then the bag with the original (but again wrong) label "green" gets now the label "red".

and the third bag with the original (but wrong) label "red" gets the "red/green" label.

if the first apple is green, then the same principle just with reversed colors applies.

7 0
1 year ago
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