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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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Find the common difference for the following sequence and fill in the missing terms. Show all calculations. 2, 13, 24, 35, __, _
Julli [10]

Step-by-step explanation:

Hey there!

Given sequences are; 2 , 13 , 24 , 35 , _ , _ .

Now,

Common difference (d) = 2nd term - 1st term. = 13-2 = 11

When we subtract 1st term from 2nd term we find 11 and when we subtract 2nd term from 3rd term we get 11. This means our common difference is 11.

Now, let's find the nrh term of the sequence.

nth term= a1 + (n-1)d ( <em>a1= 1st term, d= common</em> <em>difference</em>)

nth = 2+ (n-1) 11

= 2 + 11n - 11

= 11n - 9

Let's check if we have got nth term correct.

a1= 1*11 - 9 = 2

a2 = 2*11-9 = 13

a3 = 3*11 - 9 = 24

a4 = 4*11-9 = 35

So, we got our nth term.

Let's find remaining sequence.

a5= 5*11 - 9 = 46.

a6= 6*11 - 9 = 57.

Therefore, the remaining terms are : 46 and 57.

<em><u>Hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

7 0
3 years ago
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Find the value of each variable.<br> 105°<br> 86°<br> 106°/2°
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Answer:

z = 74 degrees

y = 115 degrees

Step-by-step explanation:

Explanation in the attachment.

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The battery standby duration (in hours) of a new model of cell phone is known to be normally distributed. Ten pieces of such new
Stells [14]

x = number of hours

want to find probability (P) x >= 13

x is N(14,1) transform to N(0,1) using z = (x - mean) / standard deviation so can look up probability using standard normal probability table.

P(x >= 13) = P( z > (13 - 14)/1) = P(z > -1) = 1 - P(z < -1) = 1 - 0.1587 = 0.8413

To convert that to percentage, multiply 100, to get 84.13%

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3 years ago
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Pls help:Find all the missing elements:
Yuliya22 [10]

Answer:

<h3>B = 48.7° , C = 61.3° , b = 12</h3>

Step-by-step explanation:

In order to find B we must first angle C

To find angle C we use the sine rule

That's

\frac{ |a|  }{ \sin(A) }  =  \frac{ |c| }{ \sin(C) }

From the question

a = 15

A = 70°

c = 14

So we have

\frac{15}{ \sin(70) }  =  \frac{14}{ \sin(C) }

\sin(C)  =  \frac{14 \sin(7 0 ) }{15}

C = \sin^{ - 1} (  \frac{14 \sin(70) }{15} )

C = 61.288

<h3>C = 61.3° to the nearest tenth</h3>

Since we've found C we can use it to find B.

Angles in a triangle add up to 180°

To find B add A and C and subtract it from 180°

That's

A + B + C = 180

B = 180 - A - C

B = 180 - 70 - 61.3

<h3>B = 48.7° to the nearest tenth</h3>

To find b we can use the sine rule

That's

\frac{ |a| }{ \sin(A) }  =  \frac{ |a| }{ \sin(B) }

\frac{15}{ \sin(70) }  =  \frac{ |b| }{ \sin(48.7) }

|b|  =  \frac{15 \sin(48.7) }{ \sin(70) }

b = 11.9921

<h3>b = 12.0 to the nearest tenth</h3>

Hope this helps you

6 0
3 years ago
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