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Katena32 [7]
3 years ago
14

What is the slope line of a and c.

Mathematics
1 answer:
faltersainse [42]3 years ago
5 0

Answer:

the amount of y over x

Step-by-step explanation:

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Drag each label to the correct location on the image.
joja [24]

Answer:

4 x (1 x 4) = 16

Step-by-step explanation:

1 x 4 = 4

4 x 4 = 16

8 0
3 years ago
30 POINTS PLEASE HELP;)
AveGali [126]

Answer:

Lorsque vous montez dans la voiture et sortez de l'aéroport, tournez à gauche. Lorsque vous arrivez à la bifurcation de la route, tournez à droite. Ensuite, vous continuerez à avancer. Vous devrez faire deux virages à gauche. À droite, vous verrez ma maison.

Step-by-step explanation: I had a similar project so I used mine as an example and I rewrote it. Hope this helps

Translation : When you first get in the car and exit the airport turn left. When you get to the fork in the road turn right. Then you will keep driving forward. You will have to make two left turns. To the right you will see my house.

Feel free to change up some words if you need to

5 0
3 years ago
Jeremy bought a new pair of jeans. The jeans regulary cost $30, but they are on sale for 10 percent off. There is a 6 percent sa
diamong [38]

Step-by-step explanation:

mp=$30

%D=10%=%D=D/M.p×100

10%=D/30$×100=100D=10

DISCOUNT =$3

S.P=$27×6÷100=$1.62+27

<h3> =$28.62</h3>

4 0
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
2, 3, 6, 8, . . . is an example of an infinite alternating series. True False
vaieri [72.5K]
Answer: False.

Explanation: An alternating series is on in which the alternate terms are negative & positive.
Example, <span>1/2 − 1/4 + 1/8 − 1/16 + ⋯  is an alternating series.

but here, in the given series - all numbers are positive . Hence, it is not an infinite alternating series.

</span>
8 0
3 years ago
Read 2 more answers
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