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Zielflug [23.3K]
3 years ago
6

What is 9-x/5=3? I kept getting 6 but the only option is -6 :/

Mathematics
2 answers:
daser333 [38]3 years ago
8 0

Answer:

Yeah the answer is -6

Step-by-step explanation:

9-x/ 5= 3

9-x = 15 (5 times three)

-x = 6 (9-6)

x= -6 ( a variable can never Be negative, but we have to give the minus to 6)

I hope this helps!

borishaifa [10]3 years ago
3 0

Step-by-step explanation:

9 -  \frac{x}{5}  = 3 \\ 9 - 5 \times  \frac{x}{5}  = 3 \times 5 \\ 9 - x = 15 \\  - 9 - x =  - 9 + 15 \\  - x = 6 \\  - x \div  - 1 = 6 \div  - 1 \\ x =  - 6

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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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Step-by-step explanation:

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