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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.

Nonagon

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

List the multiples of 40

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

Hexagon

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

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 a ---÷48=-1

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 -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