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saveliy_v [14]
2 years ago
8

Find the length s of the circular arc. (Assume r = 6 and = 128°.)

Mathematics
1 answer:
Anni [7]2 years ago
7 0

\textit{arc's length}\\\\ s=\cfrac{r\theta \pi }{180}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=6\\ \theta =128 \end{cases}\implies s=\cfrac{(6)\pi (128)}{180}\implies s=\cfrac{64\pi }{15}\implies s\approx 13.4

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A. Find the amplitude.
Feliz [49]

Answers:

  • a) Amplitude = 2
  • b) Period = pi
  • c) Vertical shift = -2, which means it has been shifted down 2 units.
  • d) Horizontal shift = 3pi/8, this shifting is to the right.
  • e) There is <u>  one  </u> cycle between 0 and 2pi.
  • f) The equation of the graph is y = 2*sin(2(x-3pi/8))-2

========================================================

Explanations:

Part (a)

The highest point is when y = 0 and the lowest point is when y = -4. The vertical distance between the peak and valley is 4 units, which cuts in half to 2. This is the amplitude. It's the vertical distance from the center to either the peak or valley.

Note: Amplitude is always positive as it measures a distance.

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

Part (b)

For x > 0, the first valley or lowest point occurs between 0 and pi/4. It appears to be the midpoint of the two values. So that would be (0+pi/4)/2 = pi/8.

The next valley occurs between pi and 5pi/4. Compute the midpoint to get (pi+5pi/4)/2 = (4pi/4+5pi/4)/2 = (9pi/4)/2 = 9pi/8

So we have the graph go from one valley x = pi/8 to the next valley over x = 9pi/8. This is a distance of pi units because 9pi/8-pi/8 = 8pi/8 = pi

The graph repeats itself every pi units, so the period is pi.

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

Part (c)

The midline is normally through y = 0, aka the x axis. However, the graph shows the midline is through y = -2. This means the graph has been shifted down 2 units.

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

Part (d)

This will depend on whether you use sine or cosine. This is entirely because cosine is a phase-shifted version of sine, and vice versa. I'll go with sine.

The parent sine function y = sin(x) goes through the origin (0,0)

However, as part (c) mentioned, we shifted the graph 2 units down. So we have y = sin(x)-2. But plugging x = 0 into this leads to the point (0,-2)

This doesn't match what the graph says. The graph shows the point (3pi/8, -2) on the red curve. The x coordinate 3pi/8 is the midpoint of pi/4 and pi/2

This must mean we need to shift the sine graph 3pi/8 units to the right.

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

Part (e)

Start at the lowest point when x = pi/8. If you start the cycle here, then it ends when x = 9pi/8. See part (b).

So far we've completed one cycle. If we start at x = 9pi/8, then the next valley or lowest point is slightly beyond or to the right of x = 2pi. This means we run out of room and we haven't completed a full cycle.

Overall, one full cycle is between 0 and 2pi.

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

Part (f)

Again I'm going to use sine instead of cosine. Refer back to part (d).

The general sine curve equation is

y = A*sin(B(x-C))+D

where

  • |A| = amplitude
  • B handles the period, specifically T = 2pi/B where T is the period. We can solve for B to get B = 2pi/T
  • C = horizontal phase shift
  • D = vertical shift, and ties together with the midline equation

In this case, we found that

  • |A| = 2
  • T = pi leads to B = 2pi/T = 2pi/pi = 2
  • C = 3pi/8
  • D = -2

So,

y = A*sin(B(x-C))+D

will update to

y = 2*sin(2(x-3pi/8))-2

which is one way to express the equation of the red curve. Optionally you can distribute the 2 through to (x-3pi/8).

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2 years ago
Tierra bought 4 3/8 yards blue ribbon and 2 1/8 yards yellow ribbon for a craft project. How much more blue ribbon than than yel
PolarNik [594]
She bought 2 1/4 more blue ribbon.

4 3/8 - 2 1/8 = 2.25
2.25= 9/4
9/4= 2 1/4

-Katie ;)
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Henry spent 1 hour 30 minutes less than Joey reading last week. Joey spent 50 minutes less than Pete. Pete spent 3 hours reading
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Answer: Henry read for a total of 40 minutes

Step-by-step explanation: 3:00 - 0:50 is 2:10 then 2:10 - 1:30= 40 minutes

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Answer:

Below, you can see the graph of the function:

f(x) = x + cos(k*x)

for different values of k, as follows:

red: k = 1

green: k = 2

orange: k = 0.

Now let's find the values of k such that our function does not have local maxima nor local minima.

First, remember that for a given function f(x), the local maxima or minima points are related to the zeros of the first derivate of f(x).

This means that if:

f'(x0) = 0.

Then x0 is a maxima, minima or an inflection point.

Then if a function is such that the f'(x) ≠ 0 , ∀x, then this function will not have local maxima nor minima.

Now we have:

f(x) = x + cos(k*x)

then:

f'(x) = 1  - k*sin(k*x)

This function will be zero when:

1 = k*sin(k*x)

1/k = sin(k*x)

now, remember that -1 ≤ sin(θ) ≤ 1

then if 1/k is smaller than -1, or larger than 1, we will not have zeros.

And this will happen if -1 < k < 1.

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2 years ago
Which of the following transforms y = x' to the graph of y = (x + 5)22
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Answer: A translation 5 units to the left... brainliest?

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