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CaHeK987 [17]
3 years ago
11

20 points and Brainliest to fastest correct answer!!

Mathematics
1 answer:
tiny-mole [99]3 years ago
6 0

Step-by-step explanation:

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Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims tha
GalinKa [24]

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.

7 0
3 years ago
If
Leno4ka [110]

Answer:

\frac{s^2-25}{(s^2+25)^2}

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given:\mathcal{L}[t \cos 5t]=(-1)F'(s) with F(s)=\mathcal{L}[\cos 5t].

Now, F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt.

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s).

Solving for F(s) on the last equation, F(s)=\frac{s}{s^2+25}, then the Laplace transform we were searching is -F'(s)=\frac{s^2-25}{(s^2+25)^2}

3 0
3 years ago
Fill in the blanks<br>a ---÷48=-1<br>​
agasfer [191]

Ans is -48.

Hope this will help u..

4 0
3 years ago
Read 2 more answers
Absolute value of +4 and -2
Greeley [361]

Answer:

4 and 2

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
How to solve <br> -3(4R-8)-36
natima [27]

Answer: -12r-12

Step-by-step explanation: There is no equals sign so you can't solve it but you can simplify it. You start out with -3(4r-8)-36. Then, you use PEMDAS. (-3×4r)+(-3×-8)-36. Then, it turns into -12r×+24-36. You just continue to simplify until you can't anymore. You end up with -12r-12.

7 0
4 years ago
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