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Hoochie [10]
3 years ago
6

The mod told me to put harder questions so here's a harder question: 2x+12= 24 what is x

Mathematics
2 answers:
MaRussiya [10]3 years ago
4 0

Answer:

x = 6

Step-by-step explanation:

2x+12=24

subtract 12 both sides

2x+12-12=24-12

2x=12

divide 2 both sides

2x/2=12/2

x = 6

hope it's helpful ❤❤❤❤

THANK YOU.

Cerrena [4.2K]3 years ago
3 0

Answer:

x=18

Step-by-step explanation:

done

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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.)

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3 years ago
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Last year's sales were $13,500. Sales have increased 47% this year. How much is the increase?
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What is the error due to using linear interpolation to estimate the value of sinxsin⁡x at x = \pi/3? your answer should have at
Bad White [126]
<h3>Answer:</h3>
  • using y = x, the error is about 0.1812
  • using y = (x -π/4 +1)/√2, the error is about 0.02620
<h3>Step-by-step explanation:</h3>

The actual value of sin(π/3) is (√3)/2 ≈ 0.86602540.

If the sine function is approximated by y=x (no error at x = 0), then the error at x=π/3 is ...

... x -sin(x) @ x=π/3

... π/3 -(√3)/2 ≈ 0.18117215 ≈ 0.1812

You know right away this is a bad approximation, because the approximate value is π/3 ≈ 1.04719755, a value greater than 1. The range of the sine function is [-1, 1] so there will be no values greater than 1.

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If the sine function is approximated by y=(x+1-π/4)/√2 (no error at x=π/4), then the error at x=π/3 is ...

... (x+1-π/4)/√2 -sin(x) @ x=π/3

... (π/12 +1)/√2 -(√3)/2 ≈ 0.026201500 ≈ 0.02620

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