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3 years ago

Which of the following could be the perimeter of the three squares below?

Mathematics

2 answers:

leonid [27]3 years ago

4 0

Answer:

Step-by-step explanation:

The length of sides would be A/4, B/4 and C/4 where A, B, and C are the perimeters of the three squares.

As the triangle is a right triangle,

c² = a² + b²

only answer A with sides 3, 4, and 5 will fulfil Pythagorean theorem.

B sides are 5, 4, and 6

C sides are 10, 20, and 30

D sides are 4, 6, and 7

stepladder [879]3 years ago

3 0

Answer:

I would say B

Step-by-step explanation:

I have done the calculations and my research matches B

XDDD

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IRISSAK [1]

3+1=4
As all parents would teach their babies simple math...take three candies and take one more. How many candies do you have now? Four

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2 years ago

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Consider the radical equation √c+22 = c + 2.

dimulka [17.4K]

First square both sides to get:

c + 22 = (c + 2)^2

c + 22 = c^2 + 4c + 4

Move the terms on the left side to the right side:

c^2 + 3c - 18 = 0

Factor to get:

(c + 6) * (c - 3) = 0.

The solutions are c = -6 and c = 3.

Check to see if these answers work by plugging them into the original equation:

c = -6:

sqrt (-6 + 22) ?= -6 + 2

But, -6 + 2 is a negative number, and you can't get a negative from a square root. So, -6 is extraneous.

c = 3:

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3 years ago

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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).

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

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

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.

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3 years ago