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eimsori [14]
2 years ago
8

Which regular polygon has a rotation of 240 degree to carry the polygon onto itself?

Mathematics
1 answer:
kirza4 [7]2 years ago
4 0

The regular polygon that has a rotation of 240 degrees to carry the polygon onto itself is the equilateral triangle.

<h3>The missing part of the question</h3>

<u>Options</u>

Equilateral triangle

Octagon

Rectangle

Pentagon

<h3>How to determine the polygon?</h3>

A polygon that can carry itself onto itself after a rotation of 240 degrees must obey the following rule:

Angle of rotation = 360/Sides

Next, we test the options:

Option (a) equilateral triangle

Angle of rotation = 360/3

Evaluate

Angle of rotation = 120

Next, we list the multiples of 120

120, 240, 360.....

See that 120 is a factor of 240.

This means that the equilateral triangle would be rotated onto itself when rotated by 240 degrees.

Read more about polygons here:

brainly.com/question/1592456

#SPJ1

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harkovskaia [24]

Answer:

First graph is 1/4

Second graph is 2

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7 0
3 years ago
Demonstrate that there is approximately 16% probability (P) of finding the ground state harmonic oscillator displaced beyond the
Misha Larkins [42]

Answer:

The probability of finding a particle in a space is proportional to the square of its absolute value.

In quantum mechanics, there are still chances of find a particle in a classically forbidden region.

That is, finding the ground state harmonic oscillator displaced beyond the classical turning points.

Since there is a chance for finding the ground state harmonic oscillator displaced beyond the classical turning points, the probability (P) will have a value and not equal to Zero( I.e 16%).

By normalization, the probability can be added to 1

This phenomenon is tunneling in quantum mechanics.

Step-by-step explanation:

The motion of a classical oscillator is confined to the region where its kinetic energy is nonnegative.

Physically, it means that a classical oscillator can never be found beyond its turning points, and its energy depends only on how far the turning points are from its equilibrium position. The energy of a classical oscillator changes in a continuous way. The lowest energy that a classical oscillator may have is zero, which corresponds to a situation where an object is at rest at its equilibrium position. The zero-energy state of a classical oscillator simply means no oscillations and no motion at all (a classical particle sitting at the bottom of the potential well.

7 0
3 years ago
Jane has earned a 91, 85, and 84 on her first three quizzes of the semester. If she hopes to have an A quiz average (90 or above
fenix001 [56]

Answer:

100

Step-by-step explanation:

91+85+84+100 is 360

Divide that by 360, and you get 90, exactly.

I hope this helps!

5 0
3 years ago
Answer this for brainilest and 30points (no links or internet answer or you will get flagged
ludmilkaskok [199]

Answer:

T=1

Step-by-step explanation: :)

7 0
3 years ago
Read 2 more answers
Any help would be appreciated ❤ ❤ Thank you
Paladinen [302]
The solid shape is a cone and a half sphere. So the equation would be (cone volume) + (1/2 sphere volume):
cone =  \frac{1}{3} \pi {r}^{2} h \\ sphere =  \frac{4}{3} \pi {r}^{3}  \\ object = ( \frac{1}{3} \pi  {r}^{2} h) +  \frac{1}{2} ( \frac{4}{3} \pi  {r}^{3} )
If the volume of the cone is 270pi and the height of the cone is 10, then:
270\pi =  \frac{1}{3} \pi {r}^{2} (10) \\ 27 =  \frac{1}{3}  {r}^{2}  \\ 80 =  {r}^{2}  \\ r = 9
The radius of the object is 9 cm. Therefore, we can plug into the formula for the total volume:
object = 270\pi +  \frac{1}{2} ( \frac{4}{3} \pi {(9)}^{3} ) \\  = 270\pi +  \frac{2}{3} (729\pi) \\  = 270\pi \: + 243\pi \\  = 513\pi
The total volume is 513 pi cm^3
7 0
3 years ago
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