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bearhunter [10]
3 years ago
11

Find the value of x.(2^3x+1)+52=180

Mathematics
2 answers:
LenaWriter [7]3 years ago
8 0
The answer is x= 127/8
Rudik [331]3 years ago
3 0

Answer:

x= 127/8

Step-by-step explanation:

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, list the first five terms of each sequence, and identify them as arithmetic or geometric.
Natali [406]

Answer:

Step-by-step explanation:

A_{n+1}=14*A_{n}

A_{1}=8

A_{1+1}=14×8=112

A_{2}=112

A_{3}=14*A_{2}=14×112=1568

A_{4}=14*A_{3}=14×1568=21952

A_{5}=14*A_{4}=14×21952=307328

7 0
3 years ago
Using the slope formula, find the slope of (1,-5) and (3,-17) pls pls help pls
inysia [295]

Answer:

-6

Step-by-step explanation:

5 0
3 years ago
A store had 5 packs of paper for $7.80. How much would it cost if you were to buy 3 packs
valentina_108 [34]
To buy 3 packs, the cost would be $4.68
4 0
3 years ago
Find the solution of the system 2x-y=0 and 4x-2=2
yanalaym [24]

Answer:

(1, 2)

Step-by-step explanation:

2x-y=0

4x-2=2

-----------

y=2x-0

y=2x

-------------

4x=2+2=4

4x=4

x=4/4=1

---------------

y=2(1)=2

x=1, y=2.

8 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
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