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USPshnik [31]
3 years ago
7

Which shape does Not stack? need help plzzzzzz its #2

Mathematics
2 answers:
Kisachek [45]3 years ago
8 0
The third one is the answer to this problem
cluponka [151]3 years ago
4 0
The shape that does not stack is the 3rd one the circle
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first one is 134 remainder 3

the second one is 211 remainder 6

Step-by-step explanation:


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How do you write 6,007,200 in expanded form
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6.007.200=6*1.000.000+7*1.000+2*100
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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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