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masha68 [24]
3 years ago
13

MATHSWATCH

Mathematics
1 answer:
Jobisdone [24]3 years ago
7 0

Answer:19683x

Step-by-step explanation:

Evaluate the exponent

27.93x

27.729x

i am pretty sure sorry if I am wrong

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Let point L be between M and N on MN. Given that MN=31, ML= h-15 and LN = 2h-8. Find LN
MrRa [10]

Since the distance of MN is 31 and we have a distance of ML of h -15 and LN of 2h – 8, therefore

ML + LN = h -15 + 2h – 8 = 31

 h = 18

substituting the value of h to the expression for LN obtaining

 

LN = 2h – 8

LN = 2(18) – 8

LN = 28

 

Therefore the length of the segment LN is 28 units

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3 years ago
Which is bigger 1 2/9or 1.2
PIT_PIT [208]
12/9 is bigger. It’s 1.333~
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What’s the expression
aev [14]

Answer: 5 + j/3

Step-by-step explanation:

Dividing the fare equally

Since they shared the j fare evenly among 3 people, you would divide the j fare by 3 to get j/3

So each person, including Angela, paid j/3 dollars before the tip

Adding the tip

Angela then added 5 dollars to this j/3 so the expression would become 5 + j/3

Final Answer

Angela paid $5 + j/3

5+\frac{j}{3}

7 0
3 years ago
Read 2 more answers
PLESASE HELP asap !!!!!!! a line passes through the point (2,6) and has a slope of -9 write the equation of the line in point sl
posledela

Answer:

the equation is (9), 1,3

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3 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
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