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love history [14]
3 years ago
11

Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims tha

t there are rotations about the center that will carry both a regular hexagon and a regular nonagon (9-sided polygon) onto itself. Determine whether each angle of rotation below can be used to support Nadia's claim. Select Yes or No for each angle of rotation
Mathematics
1 answer:
GalinKa [24]3 years ago
7 0

Answer:

\theta_1 \ n\ \theta_2 = 120, 240

Step-by-step explanation:

The question is incomplete, as the angles of rotation are not stated.

However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself

We have:

Nonagon = 9\ sides

Hexagon = 6\ sides

Divide 360 degrees by the number of sides in each angle, then find the multiples.

<u>Nonagon</u>

\theta = \frac{360}{9} =40

List the multiples of 40

\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320

<u>Hexagon</u>

\theta = \frac{360}{6} =60

List the multiples of 60

\theta_2 = 60, 120, 180, 240, 300

List out the common angles

\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320

\theta_2 = 60, 120, 180, 240, 300

\theta_1 \ n\ \theta_2 = 120, 240

This means that, only a rotation of 120, 240 will lift both shapes onto themselves, when applied to both shapes.

The other angles will only work on one of the shapes, but not both at the same time.

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If the pattern continues, how many stars will be in the Stage 15 figure?
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Answer:

There would be 45 stars in a stage 15 patteren

4 0
3 years ago
Evaluate the expression, below, when a = 2 and b = 5. <br><br> (3 + a)² • (4 – b) + √36
stellarik [79]

Answer:

-19

Step-by-step explanation:

1) substitute a with 2 and b with 5

(3 + 2)^2 * (4-5) + √36

2) solve the brackets and the square root

(5^2) * (-1) + 6

3) solve the power

25 *(-1) + 6

4) solve the multiplication

-25 + 6

5) solve the sum

- 19

6 0
2 years ago
What is the least possible value of 8x-13 when 9 is less than or equal to 3/4 + x?
Kruka [31]

The least possible value of 8x - 13 when 9 is less than or equal to 3/4 + x is;

53.

According to the question;

  • 9<=3/4 + x.

Therefore,

  • 9 -3/4 <= x

  • 8 1/4 <= x

Therefore, x >= 33/4.

The least possible value of x is therefore, 33/4

The expression 8x -3 becomes;

  • 8(33/4) -13

  • =66 - 13

  • = 53.

Therefore, the least possible value of 8x - 13 is;

53.

Read more;

brainly.com/question/18370359

6 0
3 years ago
For a history fair, a school is building a circular wooden stage that will stand that will stand 2 feet off the ground . Find th
yawa3891 [41]
To find the answer you do 19×19= 361 and do 361×3.14= 1,133.54 The area of the stage is 1,133.54
8 0
3 years ago
Read 2 more answers
Please help with imaginary numbers worksheet!
mihalych1998 [28]

Problem 5

<h3>Answer:   6i</h3>

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

Explanation:

We have these four identities

  • i^0 = 1
  • i^1 = i
  • i^2 = -1
  • i^3 = -i

Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.

We can think of it like this if you wanted

i^28 = (i^4)^7 = 1^7 = 1

Then the sqrt(-36) becomes 6i

So overall, we end up with the final answer of 6i

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

Problem 6

<h3>Answer:  -3i</h3>

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

Explanation:

We'll use the ideas mentioned in problem 5

46/4 = 11 remainder 2

i^46 = i^2 = -1

sqrt(-9) = 3i

The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i

In other words, here is one way you could write out the steps

-i^{46}*-\sqrt{-9}\\\\-i^{2}*-3i\\\\i^{2}*3i\\\\-1*3i\\\\-3i\\\\

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

Problem 7

<h3>Answer:   -1</h3>

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

Work Shown:

i^10 = i^2 because 10/4 = 2 remainder 2

i^19 = i^3 because 19/4 = 4 remainder 3

i^7 = i^3 because 7/4 = 1 remainder 3

Again, all we care about are the remainders.

i^{10}+i^{19} - i^{7}\\\\i^{2}+i^{3} - i^{3}\\\\i^{2}\\\\-1

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

Problem 8

<h3>Answer:    -1 + i</h3>

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

Work Shown:

i^22*i^6 = i^(22+6) = i^28

Earlier in problem 5, we found that i^28 = i^0 = 1

So,

i^1 - \left(i^{22}*i^{6}\right)\\\\i^1 - i^{28}\\\\i^1 - i^{0}\\\\i - 1\\\\-1+i

8 0
2 years ago
Read 2 more answers
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