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3 years ago

Which of these figures has rotational symmetry

Mathematics

1 answer:

Umnica [9.8K]3 years ago

8 0

9514 1404 393

Answer:

Step-by-step explanation:

The parallelogram has rotational symmetry of degree 2. It looks the same after rotation by 180°.

_____

<em>Additional comment</em>

When a figure only looks like itself after a full rotation of 360°, it is said to have rotational symmetry of degree 1. All of the figures here will return to their original appearance after one 360° rotation. So, we assume the intent of the question is to identify figures with a <em>rotational symmetry of degree greater than 1</em>.

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<img src="https://tex.z-dn.net/?f=solve%20for%20%22m%22%20t%3D%5Cfrac%7Bms%7D%7Bm%2Bn%7D" id="TexFormula1" title="solve for &quo

dalvyx [7]

Answer:

$\boxed{\sf \ \ \ m = -\dfrac{tn}{t-s} \ \ \ }$

Step-by-step explanation:

Hello,

let s assume that m+n is different from 0

we have this equation and we need to find m as a function of t, s, and n

$t=\dfrac{ms}{m+n}$

<=>

$(m+n)*t=ms\\\\ tm+tn=sm\\ (t-s)m = -tn\\ m = -\dfrac{tn}{t-s}$

for t-s different from 0, so t different from s

hope this helps

4 0

3 years ago

7. f(x) = Square root of quantity x plus nine. ; g(x) = 8x - 13 Find f(g(x)). (1 point) f(g(x)) = 2 Square root of quantity two

AlladinOne [14]

Answer:

f(g(x)) =

$\sqrt{8x \: - \: 13} \: + \: 9$

Step-by-step explanation:

f(x) =

$\sqrt{x} \: + \: 9$

g(x) = 8x - 13

f(g(x) =

$\sqrt{8x \: - 13} \: + \: 9$

5 0

3 years ago

Emmaline bought a new book for $12 with a coupon for 25% off, how much would it cost without the coupon

Vikentia [17]

Answer:

$16

Step-by-step explanation:

$0.75x = 12\\3x = 48\\x = 16$

8 0

3 years ago

Read 2 more answers

(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

$\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]$

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

So the characteristic polynomial is $\lambda^2-5\lambda+6=0$ .

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

$\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0$

So $\lambda=3, \lambda=2$

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If $\lambda=3$ we get the following matrix

$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

$x-2y=0$ . Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case $\lambda=2$ , using the same process, we get the vector (1,1).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that $A=PDP^{-1}$ . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

$P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]$

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

$P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]$

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

4 0

3 years ago

The length of the radius of a bicycle wheel is 13 in. Which of the following is closest to the distance the wheel will travel in

chubhunter [2.5K]

Answer:

Step-by-step explanation:

r=13in

Perimeter of a circle is 2πr

P=2×22/7×13

P=81.71in.

If it will make 3 revolution then it will have move 3 times it perimeter

3×81.71

245.14in

The distance the wheel will traveled is 245.14in

4 0

2 years ago