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Georgia [21]
2 years ago
13

Which function could be a stretch of the exponential decay function shown on the graph?

Mathematics
1 answer:
Ainat [17]2 years ago
7 0

The function that could be a stretch of the exponential decay function shown on the graph is B. f(x) = One-half(6)x.

<h3>How to illustrate the function?</h3>

From the information given, it can be seen that function A is a monotonically increasing function.

Also, it can be seen that C and D don't pass through (0, 1).

Therefore, the function could be a stretch of the exponential decay function shown on the graph is B.

Learn more about functions on:

brainly.com/question/14136983

#SPJ1

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Find the area of a sector with a central angle of 170° and a radius of 17 millimeters. Round to the nearest tenth. Question 9 op
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Answer:

428.7 mm²

Step-by-step explanation:

The area (A) of the sector is calculated as

A = area of circle × fraction of circle

   = πr² × \frac{170}{360}

   = π × 17² × \frac{17}{36}

   = 289π × \frac{17}{36}

   = \frac{289(17)\pi }{36} ≈ 428.7 mm²

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PLEASE HELP I"M ON A TIME LIMT!!
NemiM [27]

Answer: C) 1560

Step-by-step explanation:

So the first part is splitting the rectangle from the triangle so the area for the rectangle is 1440 (40 x 36) and the area for the triangle is 120 ((6x40)/ 2)

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8 0
3 years ago
For the function g(x)=3-8(1/4)^2-x
Reika [66]

Using function concepts, it is found that:

  • a) The y-intercept is y = 2.5.
  • b) The horizontal asymptote is x = 3.
  • c) The function is decreasing.
  • d) The domain is (-\infty,\infty) and the range is (-\infty,3).
  • e) The graph is given at the end of the answer.

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

The given function is:

g(x) = 3 - 8\left(\frac{1}{4}\right)^{2-x}

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

Question a:

The y-intercept is g(0), thus:

g(0) = 3 - 8\left(\frac{1}{4}\right)^{2-0} = 3 - 8\left(\frac{1}{4}\right)^{2} = 3 - \frac{8}{16} = 3 - 0.5 = 2.5

The y-intercept is y = 2.5.

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

Question b:

The horizontal asymptote is the limit of the function when x goes to infinity, if it exists.

\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2+\infty} = 3 - 8\left(\frac{1}{4}\right)^{\infty} = 3 - 8\frac{1^{\infty}}{4^{\infty}} = 3 -0 = 3

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

\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2-\infty} = 3 - 8\left(\frac{1}{4}\right)^{-\infty} = 3 - 8\times 4^{\infty} = 3 - \infty = -\infty

Thus, the horizontal asymptote is x = 3.

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

Question c:

The limit of x going to infinity of the function is negative infinity, which means that the function is decreasing.

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

Question d:

  • Exponential function has no restrictions in the domain, so it is all real values, that is (-\infty,\infty).
  • From the limits in item c, the range is: (-\infty,3)

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

The sketching of the graph is given appended at the end of this answer.

A similar problem is given at brainly.com/question/16533631

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