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4 years ago

HELP PLEASE!!

Mathematics

2 answers:

Harman [31]4 years ago

7 0

Answer: $\text{quadrilateral FSHB}\sim\text{quadrilateral KTWJ }$

Step-by-step explanation:

To polygons are said to be similar if all the corresponding angles of given polygons are equal.

In the given figure, it can be seen that in quadrilateral FSHB and quadrilateral KTWJ all the corresponding angles are equal.

∠F=∠K

∠S=∠T

∠H=∠W

∠B=∠J

Therefore, The given polygons are similar.

Hence, $\text{quadrilateral FSHB}\sim\text{quadrilateral KTWJ }$

Mariulka [41]4 years ago

3 0

Answer:

I GOT THE ANSWER OMG IT IS EASY THE ANSWER IS:

do your own homework...JUST KIDDING

it is KTWJ

Step-by-step explanation:

My teacher explained it to me, you have to match up the letters

Read more about percentage at:

brainly.com/question/843074

#SPJ1

8 0

2 years ago

D - 8 divided by 5 = -4 d =

elena55 [62]

D = 0.32

d - 8 / 5 = -4d

d - 1.6 = -4d
d + 4d = 1.6
5d = 1.6
d = 0.32

This is an exact question in my math textbook. It should be correct.

6 0

3 years ago

) a, p and d are n×n matrices. check the true statements below:

BigorU [14]

A. False. Consider the identity matrix, which is diagonalizable (it's already diagonal) but all its eigenvalues are the same (1).

b. True. Suppose $\mathbf P$ is the matrix of the eigenvectors of $\mathbf A$ , and $\mathbf D$ is the diagonal matrix of the eigenvalues of $\mathbf A$ :

$\mathbf P=\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_n\end{bmatrix}$

$\mathbf D=\begin{bmatrix}\lambda_1&&\\&\ddots&\\&&\lambda_n\end{bmatrix}$

Then

$\mathbf{AP}=\begin{bmatrix}\mathbf{Av}_1&\cdots&\mathbf{Av}_n\end{bmatrix}=\begin{bmatrix}\lambda_1\mathbf v_1&\cdots&\lambda_n\mathbf v_n\end{bmatrix}=\mathbf{PD}$

In other words, the columns of $\mathbf{AP}$ are $\mathbf{Av}_i$ , which are identically $\lambda_i\mathbf v_i$ , and these are the columns of $\mathbf{PD}$ .

c. False. A counterexample is the matrix

$\begin{bmatrix}1&1\\0&1\end{bmatrix}$

which is nonsingular, but it has only one eigenvalue.

d. False. Consider the matrix

$\begin{bmatrix}0&1\\0&0\end{bmatrix}$

with eigenvalue $\lambda=0$ and eigenvector $\begin{bmatrix}k&0\end{bmatrix}^\top$ , where $k\in\mathbb R$ . But the matrix can't be diagonalized.

7 0

3 years ago

Explain why it is useful to write linear equations in equivalent form. what does each form reveal?

defon

However, point slope equations can be awkward to use in some algebraic operations. In such cases, it may be helpful to convert the equation into a different form, the standard form. The standard form of an equation is Ax + By = C. In this kind of equation, x and y are variables and A, B, and C are integers.

3 0

3 years ago