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3 years ago

8

Can anyone help me with this one :D

1 answer:

allsm [11]3 years ago

4 0

Answer:

the answers is (10^1)^3 = 1000

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3 years ago

If a set contains 15 proper subsets then number of elements in that set

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3 years ago

Orthogonally diagonalize the matrix, giving an orthogonal matrix P and a diagonal matrix D. To save time, the eigenvalues are

alexgriva [62]

Answer:

$P=\left(\begin{array}{ccc}-\frac{2}{3}&-\frac{2}{3}&\frac{1}{3}\\\frac{1}{\sqrt{5}}&0&\frac{2}{\sqrt{5}}\\-\frac{4}{3\sqrt{5}}&\frac{\sqrt{5}}{3}&\frac{2}{3\sqrt{5}}\end{array}\right)$

Step-by-step explanation:

It is a result that a matrix $A$ is orthogonally diagonalizable if and only if $A$ is a symmetric matrix. According with the data you provided the matrix should be

$A=\left(\begin{array}{ccc}-9&-4&2\\ -4&-9&2\\2&2&-6\\\end{array}\right)$

We know that its eigenvalues are $\lambda_{1}=-14, \lambda_{2}=-5$ , where $\lambda_{2}=-5$ has multiplicity two.

So if we calculate the corresponding eigenspaces for each eigenvalue we have

$E_{\lambda_{1}=-14}=\langle(-2,-2,1)\rangle$ , $E_{\lambda_{2}=-5}=\langle(1,0,2),(-1,1,0)\rangle.$ .

With this in mind we can form the matrices $P, D$ that diagonalizes the matrix $A$ so.

$P=\left(\begin{array}{ccc}-2&-2&1\\1&0&2\\-1&1&0\\\end{array}\right)$

and

$D=\left(\begin{array}{ccc}-14&0&0\\0&-5&0\\0&0&-5\\\end{array}\right)$

Observe that the rows of $P$ are the eigenvectors corresponding to the eigen values.

Now you only need to normalize each row of $P$ dividing by its norm, as a row vector.

The matrix you have to obtain is the matrix shown below

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3 years ago

Through (-1,-5) parallel to x=2

Tamiku [17]

Paralell to x=c1 is x=c2 where c1 and c2 are constants
for x=2, passing through (-1,-5), that is (x,y)
x=-1 is the line

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3 years ago

Read 2 more answers

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Kazeer [188]

Answer:

110.7 feet cubed

Step-by-step explanation:

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3 years ago